Electrostatics 1
Electric Charge.
(1) Definition : Charge is the property associated with matter due to which it produces and experiences
electrical and magnetic effects.
(2) Origin of electric charge : It is known that every atom is electrically neutral, containing as many
electrons as the number of protons in the nucleus.
Charged particles can be created by disturbing neutrality of an atom. Loss of electrons gives positive
charge (as then
np
>
ne
) and gain of electrons gives negative charge (as then
ne
>
np
) to a particle. When an
object is negatively charged it gains electrons and therefore its mass increases negligibly. Similarly, on charging
a body with positive electricity its mass decreases. Change in mass of object is equal to
n
me
. Where,
n
is the
number of electrons transferred and
e
m
is the mass of electron
Kg
31
101.9
.
(3) Type : There exists two types of charges in nature (i) Positive charge (ii) Negative charge
Mass
M
Electron < Proton
Positively charged
+
Electron = Proton
Neutral
Mass
M
Negatively charged
–
Electron > Proton
–
–
–
–
–
+
+
+
+
2 Electrostatics
Charges with the same electrical sign repel each other, and charges with opposite electrical sign attract each
other.
(4) Unit and dimensional formula : Rate of flow of electric charge is called electric current
i.e.,
dt
dQ
i
idtdQ
, hence S.I. unit of charge is –
Ampere
sec
=
coulomb
(
C
), smaller S.I. units are
mC
,
C
,
nC
)101,101,101( 963 CnCCCCmC
. C.G.S. unit of charge is –
Stat coulomb
or
e.s.u
. Electromagnetic unit
of charge is –
ab coulomb
coulombabcoulombstatC 10
1
1031 9
. Dimensional formula
ATQ][
Note
: Benjamin Franklin was the first to assign positive and negative sign of charge.
The existence of two type of charges was discovered by Dufog.
Franklin (
i.e.
,
e.s.u
. of charge) is the smallest unit of charge while faraday is largest
(1
Faraday
= 96500
C
).
The
e.s.u.
of charge is also called stat coulomb or Franklin (
Fr
) and is related to
e.m.u
. of charge
through the relation
10
103
chargeof esu
chargeof emu
(5) Point charge : A finite size body may behave like a point charge if it produces an inverse square electric field.
For example an isolated charged sphere behave like a point charge at very large distance as well as very small
distance close to it’s surface.
(6) Properties of charge
(i) Charge is transferable : If a charged body is put in contact with an uncharged body, uncharged body
becomes charged due to transfer of electrons from one body to the other.
+
–
–
–
+
+
Electrostatics 3
(ii) Charge is always associated with mass,
i.e.
, charge can not exist without mass though mass can exist without
charge.
(iii) Charge is conserved : Charge can neither be created nor be destroyed.
e.g.
In radioactive decay the
uranium nucleus (charge
e92
) is converted into a thorium nucleus (charge
e90
) and emits an
-particle
(charge
e2
)
4
2
234
90
238
92 HeThU
. Thus the total charge is
e92
both before and after the decay.
(iv) Invariance of charge : The numerical value of an elementary charge is independent of velocity. It is proved
by the fact that an atom is neutral. The difference in masses on an electron and a proton suggests that electrons
move much faster in an atom than protons. If the charges were dependent on velocity, the neutrality of atoms
would be violated.
(v) Charge produces electric field and magnetic field : A charged particle at rest produces only electric field
in the space surrounding it. However, if the charged particle is in unaccelerated motion it produces both electric
and magnetic fields. And if the motion of charged particle is accelerated it not only produces electric and
magnetic fields but also radiates energy in the space surrounding the charge in the form of electromagnetic
waves.
(vi) Charge resides on the surface of conductor : Charge resides on the outer surface of a conductor
because like charges repel and try to get as far away as possible from one another and stay at the farthest
distance from each other which is outer surface of the conductor. This is why a solid and hollow conducting
sphere of same outer radius will hold maximum equal charge and a soap bubble expands on charging.
(vii) Charge leaks from sharp points : In case of conducting body no doubt charge resides on its outer
surface, if surface is uniform the charge distributes uniformly on the surface and for irregular surface the
v
= constant
E
and
B
but no Radiation
+
0v
E
+
v
constant
E
,
B
and Radiates energy
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
4 Electrostatics
distribution of charge,
i.e.
, charge density is not uniform. It is maximum where the radius of curvature is
minimum and vice versa.
i.e
.,
σ
/R1
. This is why charge leaks from sharp points.
(viii) Quantization of charge : When a physical quantity can have only discrete values rather than any value,
the quantity is said to be quantised. The smallest charge that can exist in nature is the charge of an electron. If
the charge of an electron (
C
19
106.1
) is taken as elementary unit
i.e.
quanta of charge the charge on any
body will be some integral multiple of
e
i.e.
,
neQ
with
....3,2,1n
Charge on a body can never be
e
3
2
,
e2.17
or
e
5
10
etc
.
Note
: Recently it has been discovered that elementary particles such as proton or neutron are
composed of quarks having charge
3/1
e
and
3/2
e
. However, as quarks do not exist in free
state, the quanta of charge is still
e
.
Quantization of charge implies that there is a maximum permissible magnitude of charge.
Comparison of Charge and Mass.
We are familiar with role of mass in gravitation, and we have just studied some features of electric charge.
We can compare the two as shown below
Charge
Mass
(1) Electric charge can be positive, negative or zero.
(1) Mass of a body is a positive quantity.
(2) Charge carried by a body does not depend upon
velocity of the body.
(2) Mass of a body increases with its velocity as
Electrostatics 5
22
0
/1 cv
m
m
where
c
is velocity of light in
vacuum,
m
is the mass of the body moving with
velocity
v
and
0
m
is rest mass of the body.
(3) Charge is quantized.
(3) The quantization of mass is yet to be established.
(4) Electric charge is always conserved.
(4) Mass is not conserved as it can be changed into
energy and vice-versa.
(5) Force between charges can be attractive or repulsive,
according as charges are unlike or like charges.
(5) The gravitational force between two masses is
always attractive.
Methods of Charging.
A body can be charged by following methods :
(1) By friction : In friction when two bodies are rubbed together, electrons are transferred from one body to
the other. As a result of this one body becomes positively charged while the other negatively charged, e.g.,
when a glass rod is rubbed with silk, the rod becomes positively charged while the silk negatively. However,
ebonite on rubbing with wool becomes negatively charged making the wool positively charged. Clouds also
become charged by friction. In charging by friction in accordance with conservation of charge, both positive and
negative charges in equal amounts appear simultaneously due to transfer of electrons from one body to the
other.
(2) By electrostatic induction : If a charged body is brought near an uncharged body, the charged body will
attract opposite charge and repel similar charge present in the uncharged body. As a result of this one side of
neutral body (closer to charged body) becomes oppositely charged while the other is similarly charged. This
process is called electrostatic induction.
–
–
–
–
–
–
–
+
+
+
+
+
+
+
+
+
+
+
+
Q
–
–
–
–
–
–
–
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Q
Q
–
–
–
–
–
–
–
+
+
+
+
+
+
+
+
+
+
+
+
Q
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
6 Electrostatics
Note
: Inducting body neither gains nor loses charge.
Induced charge can be lesser or equal to inducing charge (but never greater) and its maximum
value is given by
K
QQ' 1
1
where
Q
is the inducing charge and
K
is the dielectric constant
of the material of the uncharged body. Dielectric constant of different media are shown below
Medium
K
Vacuum / air
Water
Mica
Glass
Metal
1
80
6
5–10
Dielectric constant of an insulator can not be
For metals in electrostatics
K
and so
;QQ'
i.e. in metals induced charge is equal and
opposite to inducing charge.
(3) Charging by conduction : Take two conductors, one charged and other uncharged. Bring the
conductors in contact with each other. The charge (whether
ve
or
ve
) under its own repulsion will spread
over both the conductors. Thus the conductors will be charged with the same sign. This is called as charging by
conduction (through contact).
Uncharged
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Bodies in contact
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Both are positively charged
+
+
+
+
+
+
+
+
+
+
+
+
+
Charged
+
Electrostatics 7
Note
: A truck carrying explosives has a metal chain touching the ground, to conduct away the charge
produced by friction.
Electroscope.
It is a simple apparatus with which the presence of electric charge on a body is detected (see figure). When
metal knob is touched with a charged body, some charge is transferred to the gold leaves, which then diverges
due to repulsion. The separation gives a rough idea of the amount of charge on the body. If a charged body
brought near a charged electroscope the leaves will further diverge. If the charge on body is similar to that on
electroscope and will usually converge if opposite. If the induction effect is strong enough leaves after converging
may again diverge.
(1) Uncharged electroscope
(2) Charged electroscope
D
+
+
+
+
+
+
+
+
+
+
+
+
(A)
+
+
+
+
+
+
+
+
+
+
+
+
+
C
(C)
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
D
(B)
–
–
–
–
–
–
–
–
–
+
–
–
+
+
+
+
+
–
–
–
–
–
C
(B)
–
+
+
–
–
–
–
–
+
+
+
+
+
–
–
–
–
–
–
D
(C)
–
–
–
–
–
–
–
–
–
Charging by conduction
Charging by conduction
D
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
(A)
8 Electrostatics
Concepts
After earthing a positively charged conductor electrons flow from earth to conductor and if a negatively charged conductor is
earthed then electrons flows from conductor to earth.
When a charged spherical conductor placed inside a hollow insulated conductor and connected if through a fine conducting
wire the charge will be completely transferred from the inner conductor to the outer conductor.
Lightening-rods arrestors are made up of conductors with one of their ends earthed while the other sharp, and protects a
building from lightening either by neutralising or conducting the charge of the cloud to the ground.
With rise in temperature dielectric constant of liquid decreases.
Induction takes place only in bodies (either conducting or non-conducting) and not in particles.
If X-rays are incident on a charged electroscope, due to ionisation of air by X-rays the electroscope will get discharged and
hence its leaves will collapse. However, if the electroscope is evacuated. X-rays will cause photoelectric effect with gold and
so the leaves will further diverge if it is positively charged (or uncharged) and will converge if it is negatively charged.
If only one charge is available than by repeating the induction process, it can be used to obtain a charge many times greater
than it’s equilibrium. (High voltage generator)
+
Q
+
–
–
–
–
–
–
–
–
e
–
+
+
+
+
+
+
+
+
+
e
–
+
+
+
+
+
+
+
+
+
+
Examples based on properties of charge
Electrostatics 9
Example: 1 A soap bubble is given negative charge. Its radius will [DCE 2000; RPMT 1997; CPMT 1997; MNR 1988]
(a) Increase (b) Decrease (c) Remain unchanged (d) Fluctuate
Solution: (a) Due to repulsive force.
Example: 2 Which of the following charge is not possible
(a)
C
18
106.1
(b)
C
19
106.1
(c)
C
20
106.1
(d) None of these
Solution: (c)
,106.1 20 C
because this is
10
1
of electronic charge and hence not an integral multiple.
Example: 3 Five balls numbered 1 to 5 balls suspended using separate threads. Pair (1,2), (2,4) and (4,1) show electrostatic
attraction, while pair (2,3) and (4,5) show repulsion. Therefore ball 1 must be [NCERT 1980]
(a) Positively charged (b) Negatively charged (c) Neutral (d) Made of metal
Solution: (c) Since 1 does not enter the list of repulsion, it is just possible that it may not be having any charge. Moreover,
since ball no. 1 is being attracted by 2 and 4 both. So 2 and 4 must be similarly charged, but it is also given
that 2 and 4 also attract each other. So 2 and 4 are certainly oppositely charged.
Since 1 is attracting 2, either 1 or 2 must be neutral but since 2 is already in the list of balls repelling each
other, it necessarily has some charge, similarly 4 must have some charge. It means that though 1 is
attracting 2 and 4 it does not have any charge.
Example: 4 If the radius of a solid and hollow copper spheres are same which one can hold greater charge
[BHU 1999; KCET 1994; IIT-JEE 1974]
(a) Solid sphere (b) Hollow sphere
(c) Both will hold equal charge (d) None of these
Solution: (c) Charge resides on the surface of conductor, since both the sphere having similar surface area so they will
hold equal charge.
Example: 5 Number of electrons in one coulomb of charge will be [RPET 2001; MP PMT/PET 1998]
(a)
29
1046.5
(b)
18
1025.6
(c)
19
106.1
(d)
11
109
Solution: (b) By using
e
Q
nneQ
18
19 1025.6
106.1
1
n
Example: 6 The current produced in wire when 107 electron/sec are flowing in it [CPMT 1994]
(a) 1.6 10–26
amp
(b) 1.6 1012
amp
(c) 1.6 1026
amp
(d) 1.6 10–12
amp
Solution: (d)
amp
t
ne
t
Q
i12197106.1106.110
10 Electrostatics
Example: 7 A table-tennis ball which has been covered with a conducting paint is suspended by a silk thread so that it
hangs between two metal plates. One plate is earthed. When the other plate is connected to a high
voltage generator, the ball
(a) Is attracted to the high voltage plate and stays there
(b) Hangs without moving
(c) Swings backward and forward hitting each plate in turn
(d) None of these
Solution: (c) The table tennis ball when slightly displaced say towards the positive plate gets attracted towards the positive
plate due to induced negative charge on its near surface.
The ball touches the positive plate and itself gets positively charged by
the process of conduction from the plate connected to high voltage
generator. On getting positively charged it is repelled by the positive
plate and therefore the ball touches the other plate (earthed), which has
negative charge due to induction. On touching this plate, the positive charge of the ball gets neutralized
and in turn the ball shares negative charge of the earthed plate and is again repelled from this plate also,
and this process is repeated again and again.
Here it should be understood that since the positive plate is connected to high voltage generator, its
potential and hence its charge will always remain same, as soon as this plate gives some of its charge to
ball, excess charge flows from generator to the plate, and an equal negative charge is always induced on
the other plate.
In 1
gm
of a solid, there are 5 1021
atoms
. If one electron is removed from everyone of 0.01%
atoms of the solid, the charge gained by the solid is (given that electronic charge is 1.6 10–19
C
)
(a) + 0.08
C
(b) + 0.8
C
(c) – 0.08
C
(d) – 0.8
C
Solution: (a) To calculate charge, we will apply formula
Q
=
ne
for this, we must have number of electrons. Here,
number of electrons
%01.n
of 5 1021
i.e.
100
01.10521
n
421 10105
= 5 1017
So
Q
= 5 1017 1.6 10–19 = 8 10–2 = 0.08
C
Since electrons have been removed, charge will be positive
i.e.
Q
= + 0.08
C
Tricky example: 1
+
+
+
+
+
+
+
+
+
–
–
–
–
–
+
+
+
+
+
Electrostatics 11
Coulomb’s Law.
If two stationary and point charges
1
Q
and
2
Q
are kept at a distance
r
, then it is found that force of
attraction
or repulsion between them is
2
21
r
QQ
F
i.e
.,
221
r
QkQ
F
; (
k
= Proportionality constant)
(1) Dependence of
k
: Constant
k
depends upon system of units and medium between the two charges.
(i) Effect of units
(a) In C.G.S. for air
,1k
221
r
QQ
F
Dyne
(b) In S.I. for air
2
2
9
0
109
4
1
C
mN
k

,
221
0
.
4
1
r
QQ
F

Newton
(1
Newton
= 105
Dyne
)
Note
:
0
Absolute permittivity of air or free space =
2
2
12
1085.8 mN
C
m
Farad
. It’s
Dimension is
][ 243 ATML
0
Relates with absolute magnetic permeability (
0
) and velocity of light (
c
) according to the
following relation
00
1
c
(ii) Effect of medium
(a) When a dielectric medium is completely filled in between charges rearrangement of the charges inside
the dielectric medium takes place and the force between the same two charges decreases by a factor of
K
known as dielectric constant or specific inductive capacity (SIC) of the medium,
K
is also called relative
permittivity
r
of the medium (relative means with respect to free space).
Hence in the presence of medium
221
0
.
4
1
r
QQ
KK
F
Fair
m

Q
2
Q
1
r
Q
2
Q
1
K
r
12 Electrostatics
Here
r
K00
(permittivity of medium)
(b) If a dielectric medium (dielectric constant
K
, thickness
t
) is partially filled between the charges then
effective air separation between the charges becomes
)( Kttr
Hence force
2
21
0)(
4
1
Kttr
QQ
F

(2) Vector form of coulomb’s law : Vector form of Coulomb’s law is
,
ˆ
.. 12
221
12
321
12 r
r
qq
Kr
r
qq
KF
where
12
ˆ
r
is the unit vector from first charge to second charge along the line joining the two charges.
(3) A comparative study of fundamental forces of nature
S.No.
Force
Nature and formula
Range
Relative
strength
(i)
Force of gravitation
between two masses
Attractive
F
=
Gm
1
m
2/
r
2, obey’s
Newton’s third law of motion,
it’s a conservative force
Long range (between planets
and between electron and
proton)
1
(ii)
Electromagnetic force
(for stationary and
moving charges)
Attractive as well as repulsive,
obey’s Newton’s third law of
motion, it’s a conservative
force
Long (upto few
kelometers
)
37
10
(iii)
Nuclear force (between
nucleons)
Exact expression is not known
till date. However in some cases
empirical formula
0
/
0rr
eU
can
be utilized for nuclear potential
energy
0
U
and
0
r
are
constant.
Short (of the order of nuclear
size 10–15
m
)
1039
(strongest)
(iv)
Weak force (for
processes like
decay)
Formula not known
Short (upto 10–15
m
)
1024
Note
: Coulombs law is not valid for moving charges because moving charges produces
magnetic field also.
Coulombs law is valid at a distance greater than
.10 15 m
Q
2
Q
1
r
K
Electrostatics 13
A charge
1
Q
exert some force on a second charge
2
Q
. If third charge
3
Q
is brought near, the
force of
1
Q
exerted on
2
Q
remains unchanged.
Ratio of gravitational force and electrostatic force between (i) Two electrons is 10–43/1. (ii) Two
protons is 10–36/1 (iii) One proton and one electron 10–39/1.
Decreasing order to fundamental forces
nalGravitatioWeakneticElectromagNuclear FFFF
(4) Principle of superposition : According to the principle of super
position, total force acting on a given charge due to number of charges is
the vector sum of the individual forces acting on that charge due to all the
charges.
Consider number of charge
1
Q
,
2
Q
,
3
Q
…are applying force on a
charge
Q
Net force on
Q
will be
nnnet FFFFF 121 ..........
Concepts
Two point charges separated by a distance r in vacuum and a force F acting between them. After filling a dielectric medium
having dielectric constant K completely between the charges, force between them decreases. To maintain the force as before
separation between them changes to
Kr
. This distance known as effective air separation.
Example: 8 Two point charges
C
3
and
C
8
repel each other with a force of 40
N
. If a charge of
C
5
is added to
each of them, then the force between them will become [SCRA 1998]
(a)
N10
(b)
N10
(c)
N20
(d)
N20
Solution: (a) Initially
2
12
1083
r
kF
and Finally
2
12
1032
r
k'F
so
4
1
F
'F
N'F 10
r
1
r
2
r
3
Q
Q
1
Q
2
Q
3
Q
n – 1
Qn
Examples based on Coulomb’s law
+
Q
–
Q
14 Electrostatics
Example: 9 Two small balls having equal positive charge
Q
(coulomb) on each are suspended by two insulated string of
equal length
L
meter, from a hook fixed to a stand. The whole set up is taken in satellite into space where
there is no gravity (state of weight less ness). Then the angle between the string and tension in the string is
[IIT-JEE 1986]
(a)
2
2
0)2(
.
4
1
,180 L
Q
o

(b)
2
2
0
.
4
1
,90 L
Q

(c)
2
2
02
.
4
1
,180 L
Q

(d)
2
04
.
4
1
,180 L
QL
o

Solution: (a) In case to weight less ness following situation arises
So angle
180
and force
2
2
02
.
4
1
L
Q
F

Example: 10 Two point charges 1
C
&
C
5
are separated by a certain distance. What will be ratio of forces acting on
these two [CPMT 1979]
(a)
5:1
(b)
1:5
(c)
1:1
(d)
0
Solution: (c) Both the charges will experience same force so ratio is 1:1
Example: 11 Two charges of
C
40
and
C
20
are placed at a certain distance apart. They are touched and kept at the
same distance. The ratio of the initial to the final force between them is [MP PMT 2001]
(a)
1:8
(b)
1:4
(c) 1 : 8 (d) 1 : 1
Solution: (a) Since only magnitude of charges are changes that’s why
21qqF
1
8
1010
2040
21
21
2
1
q'q'
qq
F
F
Example: 12 A total charge
Q
is broken in two parts
1
Q
and
2
Q
and they are placed at a distance
R
from each other. The
maximum force of repulsion between them will occur, when [MP PET 1990]
(a)
R
Q
QQ
R
Q
Q 12 ,
(b)
3
2
,
412 Q
QQ
Q
Q
(c)
4
3
,
412 Q
Q
Q
Q
(d)
2
,
221 Q
Q
Q
Q
Solution: (d) Force between charges
1
Q
and
2
Q
211
221
R
QQQ
k
R
QQ
kF
For F to be maximum,
0
1
dQ
dF
i.e.
,
0
2
2
11
1
R
QQQ
k
dQ
d
or
2
,02 11 Q
QQQ
Hence
2
21 Q
QQ
L
L
+
Q
+
Q
180o
L
+
Q
L
+
Q
Electrostatics 15
Example: 13 The force between two charges 0.06
m
apart is 5
N
. If each charge is moved towards the other by 0.01
m
, then
the force between them will become [SCRA 1994]
(a) 7.20
N
(b) 11.25
N
(c) 22.50
N
(d) 45.00
N
Solution: (b) Initial separation between the charges = 0.06
m
Final separation between the charges = 0.04
m
Since
2
1
r
F
2
1
2
2
1
r
r
F
F
9
4
06.0
04.05 2
2
F
NF 25.11
2
Example: 14 Two charges equal in magnitude and opposite in polarity are placed at a certain distance apart and force
acting between them is
F
. If 75% charge of one is transferred to another, then the force between the
charges becomes
(a)
16
F
(b)
16
9F
(c)
F
(d)
F
16
15
Solution: (a)
Initially
2
2
r
Q
kF
Finally
16
4
.
2
2
F
r
Q
k
'F
Example: 15 Three equal charges each +
Q
, placed at the corners of on equilateral triangle of side a what will be the force
on any charge
0
4
1

k
[RPET 2000]
(a)
2
2
a
kQ
(b)
2
2
2
a
kQ
(c)
2
2
2
a
kQ
(d)
2
2
3
a
kQ
Solution: (d) Suppose net force is to be calculated on the charge which is kept at
A
. Two charges kept at
B
and
C
are
applying force on that particular charge, with direction as shown in the figure.
Since
2
2
a
Q
kFFF cb
So,
60cos2
22 CBCBnet FFFFF
2
2
3
3a
kQ
FFnet
Example: 16 Equal charges
Q
are placed at the four corners
A
,
B
,
C
,
D
of a square of length
a
. The magnitude of the force
on the charge at
B
will be [MP PMT 1994]
r
–
Q
+
Q
A
B
r
–
Q
/4
+
Q
/4
A
B
60o
60o
A
B
C
+
Q
+
Q
+
Q
FB
FC
16 Electrostatics
(a)
2
0
2
4
3
a
Q

(b)
2
0
2
4
4
a
Q

(c)
2
0
2
4
2
211
a
Q

(d)
2
0
2
4
2
1
2a
Q

Solution: (c) After following the guidelines mentioned above
DCADACnet FFFFFF 22
Since
2
2
a
kQ
FF CA
and
2
2
)2(a
kQ
FD
2
1
2
2
2
2
2
2
2
2
2
a
kQ
a
kQ
a
kQ
Fnet
2
221
42
0
2
a
Q

Example: 17 Two equal charges are separated by a distance
d
. A third charge placed on a perpendicular bisector at
x
distance, will experience maximum
coulomb
force when [MP PMT 2002]
(a)
2
d
x
(b)
2
d
x
(c)
22
d
x
(d)
32
d
x
Solution: (c) Suppose third charge is similar to
Q
and it is
q
So net force on it
Fnet
= 2
F
cos
Where
4
.
4
1
2
2
0d
x
Qq
F

and
4
cos 2
2d
x
x
2/3
2
2
0
2/1
2
2
2
2
0
4
4
2
4
4
.
4
1
2
d
x
Qqx
d
x
x
d
x
Qq
Fnet


for
Fnet
to be maximum
0
dx
dFnet
i.e.
0
4
4
2
2/3
2
2
0
d
x
Qqx
dx
d

or
0
4
3
4
2/5
2
22
2/3
2
2
d
xx
d
x
i.e.
22
d
x
Example: 18
ABC
is a right angle triangle in which
AB
= 3
cm
,
BC
= 4 cm and
2
ABC
. The three charges +15, +12 and –
20 e.s.u. are placed respectively on
A
,
B
and
C
. The force acting on
B
is
(a) 125
dynes
(b) 35
dynes
(c) 25
dynes
(d) Zero
Solution: (c) Net force on B
22 CAnet FFF
dyneFA20
3
1215
2
A
D
C
B
FC
FAC
FA
FD
+
Q
+
Q
+
Q
B
C
Q
Q
F
F
q
x
2
d
2
d
4/
22 dx
4/
22 dx
A
FC
FA
3
cm
4
cm
+15
esu
– 20
esu
+12
esu
B
C
22 CAnet FFF
Electrostatics 17
dyneFC15
4
2012
2
dyneFnet 25
Example: 19 Five point charges each of value +
Q
are placed on five vertices of a regular hexagon of side
L
. What is the
magnitude of the force on a point charge of value –
q
placed at the centre of the hexagon [IIT-JEE 1992]
(a)
2
2
L
Q
k
(b)
2
2
4L
Q
k
(c) Zero (d) Information is insufficient
Solution: (a) Four charges cancels the effect of each other, so the net force on the charge placed at centre due to remaining
fifth charge is
2
2
L
Q
kF
Example: 20 Two small, identical spheres having +
Q
and –
Q
charge are kept at a certain distance.
F
force acts between the
two. If in the middle of two spheres, another similar sphere having +
Q
charge is kept, then it experience a
force in magnitude and direction as [MP PET 1996]
(a) Zero having no direction (b) 8
F
towards +
Q
charge
(c) 8
F
towards –
Q
charge (d) 4
F
towards +
Q
charge
Solution: (c) Initially, force between
A
and
C
2
2
r
Q
kF
When a similar sphere
B
having charge +
Q
is
kept at the mid point
of line joining
A
and
C
, then Net force on
B
is
CAnet FFF
F
r
kQ
r
kQ
r
Q
k88
22 2
2
2
2
2
2
. (Direction is shown
in figure)
Two equal spheres are identically charged with
q
units of electricity separately. When they are
placed at a distance 3
R
from centre-to-centre where
R
is the radius of either sphere the force of
repulsion between them is
+
+
+
+
+
+
+
+
+
Tricky example: 2
r
A
C
–
Q
+
Q
B
r
/2
r
/2
+
Q
FA
FC
L
+
Q
+
Q
+
Q
+
Q
+
Q
–
Q
18 Electrostatics
(a)
2
2
0.
4
1
R
q

(b)
2
2
09
.
4
1
R
q

(c)
2
2
04
.
4
1
R
q

(d) None of these
Solution: (a) Generally students give the answer
2
2
0)3(
4
1
R
q

but it is not true. Since the charges are not
uniformly distributed, they cannot be treated as point charges and so we cannot apply coulombs
law which is a law for point charges. The actual distribution is shown in the figure above.
Electrical Field.
A positive charge or a negative charge is said to create its field around itself. If a charge
1
Q
exerts a force
on charge
2
Q
placed near it, it may be stated that since
2
Q
is in the field of
1
Q
, it experiences some force, or it
may also be said that since charge
1
Q
is inside the field of
2
Q
, it experience some force. Thus space around a
charge in which another charged particle experiences a force is said to have electrical field in it.
(1) Electric field intensity
)(E
: The electric field intensity at any point is defined as the force experienced by
a unit positive charge placed at that point.
0
q
F
E
Where
0
0q
so that presence of this charge may not affect the source charge
Q
and its electric field is
not changed, therefore expression for electric field intensity can be better written as
0
0q q
F
LimE
0
(2) Unit and Dimensional formula : It’s S.I. unit –
metercoulomb
Joule
meter
volt
coulomb
Newton
and
C.G.S. unit –
Dyne/stat coulomb
.
Dimension : [
E
] =[
13 AMLT
]
(3) Direction of electric field : Electric field (intensity)
E
is a vector quantity. Electric field due to a positive
charge is always away from the charge and that due to a negative charge is always towards the charge
(
q
0)
F
P
+
Q
Electrostatics 19
(4) Relation between electric force and electric field : In an electric field
E
a charge (
Q
) experiences a force
QEF
. If charge is positive then force is directed in the direction of field while if charge is negative force acts
on it in the opposite direction of field
(5) Super position of electric field (electric field at a point due to various charges) : The resultant electric
field at any point is equal to the vector sum of electric fields at that point due to various charges.
...
321 EEEE
The magnitude of the resultant of two electric fields is given by
cos2 21
2
2
2
1EEEEE
and the direction is given by
cos
sin
tan
21
2EE
E
(6) Electric field due to continuous distribution of charge : A system of closely spaced electric charges forms
a continuous charge distribution
Continuous charge distribution
Linear charge distribution
Surface charge distribution
Volume charge distribution
In this distribution charge distributed
on a line.
For example : charge on a wire,
charge on a ring etc. Relevant
parameter is
which is called linear
charge density
i.e.,
length
charge
In this distribution charge distributed
on the surface.
For example : Charge on a
conducting sphere, charge on a
sheet etc. Relevant parameter is
which is called surface charge
density
i.e.,
In this distribution charge distributed
in the whole volume of the body.
For example : Non conducting
charged sphere. Relevant parameter
is
which is called volume charge
density
i.e.,
volume
charge
+
+
+
+
+
+
+
+
+
R
Q
Spherical shall
+
+
+
+
+
+
+
+
+
R
+
+
+
+
+
+
Q
Non conducting
sphere
+
+
+
+
+
+
+
+
+
R
Circular charged
ring
Q
E
2
E
1
E
+
Q
E
–
Q
E
+
Q
E
F
–
Q
E
20 Electrostatics
R
Q
2
area
charge
2
4R
Q
3
3
4R
Q
To find the field of a continuous charge distribution, we divide the charge into infinitesimal charge
elements. Each infinitesimal charge element is then considered, as a point charge and electric field
dE
is
determined due to this charge at given point. The Net field at the given point is the summation of fields of all
the elements.
i.e
.,
dEE
Electric Potential.
(1) Definition : Potential at a point in a field is defined as the amount of work done in bringing a unit
positive test charge, from infinity to that point along any arbitrary path (infinity is point of zero potential).
Electric potential is a scalar quantity, it is denoted by
V
;
0
q
W
V
(2) Unit and dimensional formula : S. I. unit –
volt
Coulomb
Joule
C.G.S. unit –
Stat volt
(e.s.u.); 1
volt
300
1
Stat
volt
Dimension –
][][ 132
ATMLV
(3) Types of electric potential : According to the nature of charge potential is of two types
(i) Positive potential : Due to positive charge. (ii) Negative potential : Due to negative charge.
(4) Potential of a system of point charges : Consider
P
is a point at which net electric potential is to be
determined due to several charges. So net potential at
P
...
4
4
3
3
2
2
1
1
r
Q
k
r
Q
k
r
Q
k
r
Q
kV
In general
X
ii
i
r
kQ
V
1
Note
: At the centre of two equal and opposite charge
V
= 0
but
0E
r
1
r
2
r
3
P
– Q
1
r
4
– Q
2
+
Q
3
–
Q
4
Electrostatics 21
At the centre of the line joining two equal and similar charge
0,0 EV
(5) Electric potential due to a continuous charge distribution : The potential due to a continuous charge
distribution is the sum of potentials of all the infinitesimal charge elements in which the distribution may be
divided
i.e
.,
r
0
dQ
dVVπε4
,
(6) Graphical representation of potential : When we move from a positive charge towards an equal
negative charge along the line joining the two then initially potential decreases in magnitude and at centre
become zero, but this potential is throughout positive because when we are nearer to positive charge, overall
potential must be positive. When we move from centre towards the negative charge then though potential
remain always negative but increases in magnitude fig. (A). As one move from one charge to other when both
charges are like, the potential first decreases, at centre become minimum and then increases Fig. (B).
(7) Potential difference : In an electric field potential difference between two points
A
and
B
is defined as
equal to the amount of work done (by external agent) in moving a unit positive charge from point
A
to point
B.
i.e.,
0
q
W
VV AB
in general
VQW .
;
V
Potential difference through which charge
Q
moves.
Electric Field and Potential Due to Various Charge Distribution.
(1) Point charge : Electric field and potential at point
P
due to a point charge
Q
is
+
q
–
q
V
x
Y
X
O
(A)
v
x
Y
X
O
(B)
+
q
+
q
22 Electrostatics
r
r
Q
kE ˆ
or 2
2
r
Q
kE
0
4
1

k
,
r
Q
kV
Note
: Electric field intensity and electric potential due to a point charge
q
, at a distance
t
1 +
t
2 where
t
1 is thickness of medium of dielectric constant
K
1 and
t
2 is thickness of medium of dielectric constant
K
2 are :
2
221
0)(
4
1
KtKt
Q
E
1
πε
;
)
221
0
4
1
KtK(t
Q
V
1
πε
(2) Line charge
(i) Straight conductor : Electric field and potential due to a charged straight conducting wire of length
l
and
charge density
(a) Electric field :
)sin(sin
r
k
Ex
and
)cos(cos
r
k
Ey
If
=
;
sin
2
r
k
Ex
and
Ey
= 0
If
l
i.e.
=
=
2
;
r
k
Ex
2
and
Ey
= 0 so
r
Enet 0
2

If
= 0,
2
;
r
k
EE yx
||||
so
r
k
EEE yxnet
2
22
(b) Potential :
1
1
log
222
22
0lr
lr
Ve

for infinitely long conductor
crV e
log
20

r
Q
P
+
+
+
+
l
+
P
r
Ey
Ex
Electrostatics
17
(ii) Charged circular ring : Suppose we have a charged circular ring of radius R and charge
Q. On it’s axis electric field and potential is to be determined, at a point ‘x’ away from the
centre of the ring.
(a) Electric field : Consider an element carrying charge dQ . It’s electric field
22 xR
KdQ
dE
directed as shown. It’s component along the axis is
cosdE and perpendicular to the axis is
sindE . By symmetry 0sin
dE , hence 212222 )(
.
)(
cos xR
x
xR
kdQ
dEE
23
22 xR
kQx
E
directed away from the centre if Q is positive
(b) Potential : 22
0
.
4
1
Rx
Q
V
Note
: At centre x = 0 so Ecentre= 0 and
R
kQ
Vcentre
At a point on the axis such that x >> R 2
x
kQ
E and
x
kQ
V
At a point on the axis if
2
R
x , 2
0
max 36 a
Q
E
(3) Surface charge :
(i) Infinite sheet of charge : Electric field and potential at a point P as shown
)(
20
o
rEE
and C
r
V
0
2
(ii) Electric field due to two parallel plane sheet of charge : Consider two large,
uniformly charged parallel. Plates A and B, having
surface charge densities are A
and B
respectively. Suppose net electric field at points
P, Q and R is to be calculated.
At P, )(
2
1
)(
0
BABAP EEE
x
+
O
2
R
2
R
x
E
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
r
P
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
A
R
P
E
B
E
A
E
B
E
A
E
A
E
B
B
Q
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
dE
sin
dE
cos
Ed
R
P
x
dQ
18
Electrostatics
At Q, )(
2
1
)(
0
BABAQ EEE
; At R, )(
2
1
)(
0
BABAR EEE
Note
: If
A and
B then 0,,0
0
RQp EEE
. Thus in case of two infinite
plane sheets of charges having equal and opposite surface
charge densities, the field is non-zero only in the space between
the two sheets and is independent of the distance between them
i.e., field is uniform in this region. It should be noted that this
result will hold good for finite plane sheet also, if they are held
at a distance much smaller then the dimensions of sheets i.e.,
parallel plate capacitor.
(iii) Conducting sheet of charge :
0
E
C
r
V
0
(iv) Charged conducting sphere : If charge on a conducting sphere of radius R is Q as
shown in figure then electric field and potential in different situation are –
(a) Out side the sphere : P is a point outside the sphere at a distance r from the centre at
which electric field and potential is to be determined.
Electric field at P
2
0
2
2
0
.
4
1
r
R
r
Q
Eout
and r
R
r
Q
Vout
0
2
0
.
4
1
2
4R
AQ
(b) At the surface of sphere : At surface
R
r
So,
0
2
0
.
4
1
R
Q
Es and
00
.
4
1
R
R
Q
Vs
(c) Inside the sphere : Inside the conducting charge sphere electric field is zero and
potential remains constant every where and equals to the potential at the surface.
+
+
+
+
+
+
+
+
+
+
r
P
+
Q
R
+
+
+
+
+
+
+
+
+
+
+
+
+
R
+
Q
+
+
+
+
+
+
+
+
+
+
+
+
+
Hollo
w
Solid
E
=
/
0
+
+
+
+
+
+
–
–
–
–
–
–
Electrostatics
19
0
in
E and in
V= constant s
V
Note
: Graphical variation of electric field and potential of a charged spherical conductor
with distance
(4) Volume charge (charged non-conducting sphere) :
Charge given to a non conducting spheres spreads uniformly throughout it’s volume.
(i) Outside the sphere at P
2
0
.
4
1
r
Q
Eout
and r
Q
Vout .
4
1
0
by using
3
3
4R
Q
2
0
3
3r
R
Eout
and r
R
Vout
0
3
3
(ii) At the surface of sphere : At surface
R
r
0
2
03
.
4
1
R
R
Q
Es and
0
2
03
.
4
1
R
R
Q
Vs
(iii) Inside the sphere : At a distance r from the centre
3
0
.
4
1
R
Qr
Ein
0
3
r
rEin and
0
22
3
22
06
)3(
2
]3[
4
1
rR
R
rRQ
Vin
Note
: At centre
0
r
So, s
V
R
Q
V2
3
.
4
1
2
3
0
centre
i.e.,
outsurfacecentre VVV
Graphical variation of electric field and potential with distance
+
Q
r
P
R
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
O
O
R
r
r
out
V
1
V
S
r =R
V
-
r
graph
O
E
in
=
0
E
O
R
r
2
1
r
Eout
E
-
r
graph
E
-
r
graph
E
2
1
r
Eout
E
in
r
R
O
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
V
-
r
graph
R
O
+
+
+
+
+
+
+
+
+
+
+
+
+
+
r
Vout
1
V
C
V
S
+
20
Electrostatics
(5) Electric field and potential in some other cases
(i) Uniformly charged semicircular ring : length
charge
At centre :
2
0
2
2
2
R
Q
R
K
E
R
Q
R
KQ
V
0
4
(iii) Charged cylinder of infinite length
(a) Conducting (b) Non-conducting
For both type of cylindrical charge distribution r
Eout
0
2
, and R
Esuface
0
2
but for
conducting 0
in
Eand for non-conducting 2
0
in 2R
r
E
. (we can also write formulae in form
of
i.e., r
R
E
0
2
out 2
etc.)
(ii) Hemispherical charged body :
At centre O,
0
4
E
0
2
R
V
(iv) Uniformly charged disc
At a distance x from centre O on it’s axis
R
x
O
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
O
+
Q
+
R
+
+
+
+
+
+
+
+
+
+
r
P
R
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
r
P
Electrostatics
21
22
0
1
2Rx
x
E
xRxV 22
0
2
Note
: Total charge on disc Q =

R2
If x 0,
0
2
–
~
E i.e. for points situated near the disc, it behaves as an infinite
sheet of charge.
Concepts
No point charge produces electric field at it’s own position.
Since charge given to a conductor resides on it’s surface hence electric field inside it is zero.
The electric field on the surface of a conductor is directly proportional to the surface charge density at that
point i.e,
E
Two charged spheres having radii 1
r and 2
r charge densities 1
and 2
respectively, then the ratio of
electric field on their surfaces will be 2
1
2
2
2
1
2
1
r
r
E
E
2
4r
Q
In air if intensity of electric field exceeds the value CN /103 6
air ionizes.
A small ball is suspended in a uniform electric field with the help of an insulated thread. If a high energy x–
ray bean falls on the ball, x-rays knock out electrons from the ball so the ball is positively charged and
therefore the ball is deflected in the direction of electric field.
Electric field is always directed from higher potential to lower potential.
A positive charge if left free in electric field always moves from higher potential to lower potential while a
negative charge moves from lower potential to higher potential.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
E
= 0
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
E
= 0
+
F
=
QE
X
–
R
ay
E
+
q
+
Q
+
+
Q
+
+
+
+
+
+
+
–
–
–
–
–
–
22
Electrostatics
The practical zero of electric potential is taken as the potential of earth and theoretical zero is taken at
infinity.
An electric potential exists at a point in a region where the electric field is zero and it’s vice versa.
A point charge +Q lying inside a closed conducting shell does not exert force another point charge q placed
outside the shell as shown in figure
Actually the point charge +Q is unable to exert force on the charge +q because it can not produce electric field
at the position of +q. All the field lines emerging from the point charge +Q terminate inside as these lines
cannot penetrate the conducting medium (properties of lines of force).
The charge q however experiences a force not because of charge +Q but due to charge induced on the outer surface of
the shell.
Example: 21 A half ring of radius R has a charge of
per unit length. The electric field at the centre is
0
4
1

k
[CPMT 2000; CBSE PMT 2000; REE 1999]
(a) Zero (b)
R
k
(c)
R
k
2 (d)
R
k

Solution: (c)
Rddl
Charge on .
Rddl
Field at C due to dE
R
Rd
kdl 2
.
We need to consider only the component
cosdE , as the component
sindE will cancel out
because of the field at C due to the symmetrical element dl
,
The total field at C is
2
0
cos2
dE
R
kd
R
k
2cos2
2
0
2
0
2R
Q

Example: 22 What is the magnitude of a point charge due to which the electric field 30 cm away has the
magnitude 2 newton/coulomb ]1094/1[ 29
0Nm

(a)
coulomb
11
10
2
(b)
coulomb
11
10
3
(c)
coulomb
11
10
5
(d)
coulomb
11
10
9
Solution: (a) By using 2
0
.
4
1
r
Q
E

;
2
2
9
1030
1092
Q
CQ 11
102
–
e
+
e
E
0
V
= 0
Examples based on electric field and electric
potential
d
dl
dl
dE
C
Electrostatics
23
Example: 23 Two point charges Q and – 3Q are placed at some distance apart. If the electric field at the
location of Q is E, then at the locality of – 3Q, it is
(a)
E
(b) E/3 (c)
E
3
(d) – E/3
Solution: (b) Let the charge Q and – 3Q be placed respectively at A and B at a distance x
Now we will determine the magnitude and direction to the field produced by charge – 3Q at
A, this is E as mentioned in the Example.
2
3
x
Q
E (along AB directed towards negative charge)
Now field at location of – 3Q i.e. field at B due to charge Q will be 3
2
E
x
Q
E' (along AB
directed away from positive charge)
Example: 24 Two charged spheres of radius 1
R and 2
R respectively are charged and joined by a wire.
The ratio of electric field of the spheres is
(a)
2
1
R
R (b)
1
2
R
R (c) 2
2
2
1
R
R (d) 2
1
2
2
R
R
Solution: (b) After connection their potential becomes equal i.e.,
2
2
1
1.
.R
Qk
R
Q
k;
2
1
2
1
R
R
Q
Q
Ratio of electric field .
1
2
2
1
2
2
1
2
1
R
R
R
R
Q
Q
E
E
Example: 25 The number of electrons to be put on a spherical conductor of radius 0.1m to produce an
electric field of 0.036 N/C just above its surface is
(a) 5
107.2 (b) 5
10
6
.
2
(c) 5
10
5
.
2
(d) 5
10
4
.
2
Solution: (c) By using 2
R
Q
kE , where R = radius of sphere so 0.036 =
2
9
1.0
109 ne
5
10
5
.
2
n
Example: 26 Eight equal charges each +Q are kept at the corners of a cube. Net electric field at the centre
will be
0
4
1

k
(a) 2
r
kQ (b) 2
8
r
kQ (c) 2
2
r
kQ (d) Zero
Solution: (d) Due to the symmetry of charge. Net Electric field at centre is zero.
Note
:
Example: 27 q, 2q, 3q and 4q charges are placed at the four corners A, B, C and D of a square. The field
at the centre O of the square has the direction along.
A
B
x
Q
–
3
Q
+q
a
a
a
+q
+q
E
=0
Equilateral
triangle
a
a
a
a
q
q
q
q
E
=
0
Square
2
q
q
O
A
B
24
Electrostatics
(a) AB (b) CB (c) AC (d) BD
Solution: (b) By making the direction of electric field due to all charges at centre. Net electric field has the
direction along CB
Example: 28 Equal charges Q are placed at the vertices A and B of an equilateral triangle ABC of side a.
The magnitude of electric field at the point A is
(a) 2
0
4a
Q

(b) 2
0
4
2
a
Q

(c) 2
0
4
3
a
Q

(d) 2
0
2a
Q

Solution: (c) As shown in figure Net electric field at A
60cos2
22
CBCB EEEEE
2
0
.
4
1
a
Q
EE CB

So, 2
0
4
3
a
Q
E

Example: 29 Four charges are placed on corners of a square as shown in figure having side of 5 cm. If Q
is one micro coulomb, then electric field intensity at centre will be
(a) CN /1002.1 7
upwards
(b) CN /1004.2 7
downwards
(c) CN /1004.2 7
upwards
(d) CN /1002.1 7
downwards
Solution: (a) |||
|
AC EE so resultant of AC EE & is ACCA EEE directed toward Q
Also |||
|
DB EE so resultant of B
Eand D
E i.e.
DBBD EEE directed toward – 2Q charge hence Net electric field at centre is
22
BDCA EEE .… (i)
By proper calculations CNE A/1072.0
10
2
5
10
109|
|
7
2
2
6
9
60
o
A
B
C
+
Q
+
Q
F
B
F
C
E
a
a
a
Q
–
2
Q
+ 2
Q
–
Q
–
2
Q
Q
+2
Q
–
Q
O
A
B
C
D
E
CA
E
C
E
B
E
A
E
D
E
BD
E
net
Electrostatics
25
CNEB/1044.1
10
2
5
102
109|
|
7
2
2
6
9
; CNEC/1044.1
10
2
5
102
109|
|
7
2
2
6
9
CNED/1072.0
10
2
5
10
109|
|
7
2
2
6
9
; So, N/C.||E||E|
|E
ACCA
7
10720
and ./1072.0|||||
|
7CNEEE DBBD Hence from equation – (i)
CNE /1002.1 7
upwards
Example: 30 Infinite charges are lying at x = 1, 2, 4, 8…meter on X-axis and the value of each charge is
Q. The value of intensity of electric field and potential at point x = 0 due to these charges
will be respectively
(a) Q
9
1012 N/C, 1.8 104 V (b) Zero, 1.2 104V
(c) Q
9
106 N/C, 9 103 V (d) Q
9
104 N/C , 6 103
V
Solution: (a) By the superposition, Net electric field at origin
...
8
1
4
1
2
1
1
1
2222
kQE
...
64
1
16
1
4
1
1kQE
...
64
1
16
1
4
1
1 is an infinite geometrical progression it’s sum can be obtained by using
the formula
r
a
S
1
; Where a = First term, r = Common ratio.
Here
1
a
and
4
1
r so, 3
4
4/11
1
.....
64
1
16
1
4
1
1
.
Hence CNQQE /1012
3
4
109 99
Electric potential at origin
.......
8
101
4
101
2
101
1
101
4
16666
0

V
2
1
1
1
109............
8
1
4
1
2
1
110109 369 volt
4
108.1
x
= 0
x
= 1
x
= 2
x
= 4
x
= 8
26
Electrostatics
Note
: In the arrangement shown in figure +Q and – Q are alternatively
and equally spaced from each other, the net potential at the origin O is
x
Q
Ve2log
.
4
1
0

[IIT 1998]
Example: 31 Potential at a point x-distance from the centre inside the conducting sphere of radius R and
charged with charge Q is [MP PMT 2001]
(a)
R
Q (b)
x
Q (c) 2
x
Q (d) xQ
Solution: (a) Potential inside the conductor is constant.
Example: 32 The electric potential at the surface of an atomic nucleus (Z = 50) of radius
V
5
10
9
is
(a) 80 V (b) V
6
108 (c) 9 V (d) V
5
109
Solution: (b) V
r
ne
V6
15
19
99 108
109
106.150
109109
Example: 33 Eight charges having the valves as shown are arranged symmetrically on a circle of radius
0.4m in air. Potential at centre O will be
(a) volt
4
1063 (b) volt
10
1063 (c)
volt
6
10
63
(d) Zero
Solution: (a) Due to the principle of superposition potential at O
voltV 4
6
9
6
0
1063
4.0
1028
109
4.0
1028
4
1

Example: 34 As shown in the figure, charges +q and –q are placed at the vertices B and C of an isosceles
triangle. The potential at the vertex A is
+
Q
–
Q
+
Q
–
Q
x
2
x
3
x
4
x
O
+5
C
+11
C
–
5
C
+7
C
–
5
C
+7
C
+15
C
–
7
C
O
A
B
C
–
q
+
q
a
b
b
Electrostatics
27
(a)
22
0
2
.
4
1
ba
q

(b)
22
0
.
4
1
ba
q

(c)
22
0
)(
.
4
1
ba
q

(d) Zero
Solution: (d) Potential at A = Potential due to (+q) charge + Potential due to (– q) charge
0
)(
4
1
.
4
1
22
0
22
0
ba
q
ba
q

Example: 35 A conducting sphere of radius R is given a charge Q. consider three points B at the surface,
A at centre and C at a distance R/2 from the centre. The electric potential at these points
are such that [DCE 1994]
(a) VA = VB = VC (b) VA = VB VC (c) VA VB VC (d) VA VB = VC
Solution: (a) Potential inside a conductor is always constant and equal to the potential at the surface.
Example: 36 Equal charges of 9
10
3
10
coulomb are lying on the corners of a square of side 8 cm. The
electric potential at the point of intersection of the diagonals will be
(a) 900 V (b) V2900 (c) V2150 (d) V21500
Solution: (d) Potential at the centre O
2/
.
4
1
4
0a
Q
V

given CQ 9
10
3
10
mcma 2
1088
2
108
10
3
10
1095 2
9
9
Vvolt21500
A point charge Q is placed outside a hollow spherical conductor of radius R
, at a distance
(r > R) from its centre C. The field at C due to the induced charges on the conductor is
0
4
1

K
(a) Zero (b)
2
Rr
Q
K
(c) 2
r
Q
K directed towards Q (d) 2
r
Q
K
directed
away from Q
Solution: (c) A according to the figure shown below. The total field at C must be zero. The field at C
due to the point charge is 2
r
Q
KE towards left. The field at C
due to the induced
charges must be 2
r
KQ towards right i.e. directed towards Q.
A point charge
q
is placed at a distance of
r
from the centre of an uncharged
+
–
–
–
–
–
–
+
+
+
+
+
+
r
C
R
+
Q
Tricky example: 3
Tricky example: 4
O
Q
Q
Q
Q
a
2a
28
Electrostatics
conducting sphere of radius
R
(<
r
). The potential at any point on th
e sphere is
(a) Zero (b) r
q
.
4
1
0

(c) 2
0
.
4
1
r
qR

(d) R
qr 2
0
.
4
1

Solution: (c) Since, potential V
is same for all points of the sphere. Therefore, we can calculate its
value at the centre of the sphere.
'.
4
1
0
V
r
q
V

; where V
= potential at centre due to induced charge = 0 (because
net induced charge will be zero) r
q
V.
4
1
0

.
Potential Due to Concentric Spheres.
To find potential at a point due to concentric sphere following guideline are to be
considered
Guideline 1: Identity the point (P) at which potential is to be determined.
Guideline 2: Start from inner most sphere, you should know where point (P) lies w.r.t.
concerning sphere/shell (i.e. outside, at surface or inside)
Guideline 3: Then find the potential at the point (P) due to inner most sphere and then due to
next and so on.
Guideline 4: Using the principle of superposition find net potential at required
shell/sphere.
Standard cases
Case (i) : If two concentric conducting shells of radii r1 and r2(r2 > r1) carrying uniformly
distributed charges Q1 and Q2 respectively. What will be the potential of each shell
To find the solution following guidelines are to be taken.
Here after following the above guideline potential at the surface of inner shell is
2
2
01
1
0
1.
4
1
.
4
1
r
Q
r
Q
V

and potential at the surface of outer shell
2
2
02
1
0
2.
4
1
.
4
1
r
Q
r
Q
V

Q
1
r
1
r
2
Q
2
Electrostatics
29
Case (ii) : The figure shows three conducting concentric shell of radii a, b and c (a < b < c)
having charges Qa, Qb and Qc respectively what will be the potential of each shell
After following the guidelines discussed above
Potential at A;
c
Q
b
Q
a
Q
Vcba
A
0
4
1

Potential at B;
c
Q
b
Q
b
Q
Vcba
B
0
4
1

Potential at C;
c
Q
c
Q
c
Q
Vcba
C
0
4
1

Case (iii) : Th
e figure shows two concentric spheres having radii
r
1
and
r
2
respectively (
r
2
>
r
1
). If
charge on inner sphere is +Q and outer sphere is earthed then determine.
(a) The charge on the outer sphere
(b) Potential of the inner sphere
(i) Potential at the surface of outer sphere 0.
4
1
.
4
1
2020
2 r
Q'
r
Q
V

QQ'
(ii) Potential of the inner sphere
2010
1
)(
4
1
.
4
1
r
Q
r
Q
V

210
11
4rr
Q

Case (iv) : In the case III if outer sphere is given a charge +
Q
and inner sphere is earthed then
(a) What will be the charge on the inner sphere
(b) What will be the potential of the outer sphere
(i) In this case potential at the surface of inner sphere is zero, so if Q' is the charge induced on
inner sphere
then 0
4
1
210
1
r
Q
r
Q'
V

i.e., Q
r
r
Q'
2
1
(Charge on inner sphere is less than that of the outer sphere.)
(ii) Potential at the surface of outer sphere
2020
2.
4
1
.
4
1
r
Q
r
Q'
V

Q
r
r
Q
r
V
2
1
20
24
1

2
1
20
1
4r
r
r
Q

Example: 37 A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10
volts. The potential at the centre of the sphere is
(a) Zero (b) 10 V
(c) Same as at a point 5 cm away from the surface (d) Same as at a point 25
cm away from the surface
Solution: (b) Inside the conductors potential remains same and it is equal to the potential of surface, so
here potential at the centre of sphere will be 10 V
Q
+
Q
r
1
r
2
Q
c
Q
b
Q
a
c
b
a
B
A
r
2
r
1
+
Q
Examples based on concentric
spheres
30
Electrostatics
Example: 38 A sphere of 4 cm radius is suspended within a hollow sphere of 6 cm radius. The inner
sphere is charged to a potential 3 e.s.u. When the outer sphere is earthed. The charge on
the inner sphere is [MP PMT 1991]
(a) 54 e.s.u. (b)
4
1e.s.u. (c) 30 e.s.u. (d) 36 e.s.u.
Solution: (d) Let charge on inner sphere be +Q. charge induced on the inner surface of outer sphere will
be –Q.
So potential at the surface of inner sphere (in CGS)
6
4
3QQ
36Q e.s.u.
Example: 39 A charge Q is distributed over two concentric hollow spheres of radii r and )( rR such that
the surface densities are equal. The potential at the common centre is
(a) )(4
)(
0
22
rR
rRQ

(b)
r
R
Q
(c) Zero (d) )(4
)(
22
0rR
rRQ

Solution: (d) If 1
q and 2
q are the charges on spheres of radius r and R respectively, in accordance with
conservation of charge
21 qqQ ….(i)
and according to the given problem 21
i.e., 2
2
2
1
44 R
q
r
q
2
2
2
1
R
r
q
q …. (ii)
So equation (i) and (ii) gives )( 22
2
1rR
Qr
q
and )( 22
2
2rR
QR
q
Potential at common centre
R
q
r
q
V21
0
4
1

)(
)(
.
4
1
)()(
4
1
22
0
2222
0rR
rRQ
rR
QR
rR
Qr

Example: 40 A solid conducting sphere having a charge Q is surrounded by an uncharged concentric
conducting hollow spherical shell. Let the potential difference between the surface of the
solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given
a charge of – 3Q, the new potential difference between the two surfaces is
(a) V (b) 2V (c) 4V (d) –2V
Solution: (a) If a and b are radii of spheres and spherical shell respectively, potential at their surfaces
will be
a
Q
V.
4
1
0
sphere

and b
Q
V.
4
1
0
shell

and so according to the given problem.
shellsphere VVV
ba
Q11
40

…. (i)
Now when the shell is given a charge –3Q the potential at its surface and also inside will
change by
b
Q
V3
4
1
0
0

4
cm
6
cm
+
Q
q
1
q
2
r
R
+
+
+
+
+
+
+
+
+
+
+
+
+
Q
a
b
Sphere
Electrostatics
31
So that now
b
Q
a
Q
V3
4
1
0
sphere

and
b
Q
b
Q
V3
4
1
0
shell

hence
V
ba
Q
VV
11
40
shellsphere

Example: 41 Three concentric metallic spheres A, B and C have radii a, b and c )( cba and surface
charge densities on them are
,
and
respectively. The valves of A
V and B
V will be
(a)
cb
b
a
cba
2
00
),(
(b)
c
a
cba
2
),(
(c)
cb
c
a
cba
2
00 ),(
(d)
c
c
b
c
a22
0
and )(
0
cba
Solution: (a) Suppose charges on A, B and C are ba qq , and c
q
Respectively, so 2
24
4aq
a
q
a
a
A
, 2
24
4bq
b
q
b
b
B
and 2
24
4cq
c
q
c
c
C
Potential at the surface of A
ininsurface )()()( CBAA VVVV
c
q
b
q
a
qcba
0
4
1

c
c
b
b
a
a222
0
44)(4
4
1

]
0
cbaVA
Potential at the surface of B
c
q
b
q
b
q
VVVV cba
CBAB
0
insurfaceout 4
1
)()()(

c
c
b
b
b
a222
0
444
4
1

cb
b
a2
0
Electric Lines of Force.
(1) Definition : The electric field in a region is represented by continuous lines (also called
lines of force). Field line is an imaginary line along which a positive test charge will move if left
free.
Electric lines of force due to an isolated positive charge, isolated negative charge and due
to a pair of charge are shown below
+
–
+
–
N
+
+
A
B
C
a
b
c
32
Electrostatics
(2) Properties of electric lines of force
(i) Electric field lines come out of positive charge and go into the negative charge.
(ii) Tangent to the field line at any point gives the direction of the field at that point.
(iii) Field lines never cross each other.
(iv) Field lines are always normal to conducting surface.
(v) Field lines do not exist inside a conductor.
(vi) The electric field lines never form closed loops. (While magnetic lines of forces form
closed loop)
(vii) The number of lines originating or terminating on a charge is proportional to the
magnitude of charge. In the following figure electric lines of force are originating from A and
terminating at B hence QA is positive while QB is negative, also number of electric lines at force
linked with QA are more than those linked with QB hence |Q||Q
|
BA
A
B
E
A
E
B
(A)
+
+
+
+
+
+
–
–
–
–
–
–
(B)
A
B
Q
A
Q
B
+
–
N
S
Electrostatics
33
(viii) Number of lines of force per unit area normal to the area at a point represents
magnitude of intensity (concept of electric flux i.e., EA
)
(ix) If the lines of forces are equidistant and parallel straight lines the field is uniform and
if either lines of force are not equidistant or straight line or both the field will be non uniform,
also the density of field lines is proportional to the strength of the electric field. For example
see the following figures.
(3) Electrostatic shielding : Electrostatic shielding/screening is the phenomenon of
protecting a certain region of space from external electric field.
Sensitive instruments and appliances are affected seriously with
strong external electrostatic fields. Their working suffers and
they may start misbehaving under the effect of unwanted fields.
The electrostatic shielding can be achieved by protecting
and enclosing the sensitive instruments inside a hollow
conductor because inside hollow conductors, electric fields is
zero.
(i) It is for this reason that it is safer to sit in a car or a bus during lightening rather than
to stand under a tree or on the open ground.
(ii) A high voltage generator is usually enclosed in such a cage which is earthen. This would
prevent the electrostatic field of the generator from spreading out of the cage.
(iii) An earthed conductor also acts as a screen against the electric field. When conductor is
not earthed field of the charged body C due to
electrostatic induction continues beyond AB. If
AB is earthed, induced positive charge
neutralizes and the field in the region beyond
AB disappears.
E
X
Y
(A
)
E
X
=
E
X
Y
(B
)
E
X
>
E
+
+
+
+
+
+
+
+
–
–
–
–
–
–
–
–
E
E
E
E
E
= 0
q
= 0
Hollow space
Conductor
C
–
A
B
+
+
+
+
+
–
–
–
–
–
C
A
B
–
–
–
–
–
34
Electrostatics
Equipotential Surface or Lines.
If every point of a surface is at same potential, then it is said to be an equipotential surface
or
for a given charge distribution, locus of all points having same potential is called
“equipotential surface” regarding equipotential surface following points should keep in mind :
(1) The density of the equipotential lines gives an idea about the magnitude of electric field.
Higher the density larger the field strength.
(2) The direction of electric field is perpendicular to the equipotential surfaces or lines.
(3) The equipotential surfaces produced by a point charge or a spherically charge
distribution are a family of concentric spheres.
(4) For a uniform electric field, the equipotential surfaces are a family of plane
perpendicular to the field lines.
(5) A metallic surface of any shape is an equipotential surface e.g. When a charge is given
to a metallic surface, it distributes itself in a manner such that its every point comes at same
potential even if the object is of irregular shape and has sharp points on it.
If it is not so, that is say if the sharp points are at higher potential then due to potential
difference between these points connected through metallic portion, charge will flow from
V
=
V
2
V
=
V
1
V
1
V
2
V
3
V
4
V
5
Equipotentia
l surface
V
1
>
V
2
>
V
3
>
V
4
>
V
5
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
V =
const.
O
Metallic charged
sphere
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
V
=
const.
Charged metallic body of irregular
shape
Electrostatics
35
points of higher potential to points of lower potential until the potential of all points become
same.
(6) Equipotential surfaces can never cross each other
(7) Equipotential surface for pair of charges
Concepts
Unit field i.e. 1N/C is defined arbitrarily as corresponding to unit density of lines of force.
Number of lines originating from a unit charge is
0
1
It is a common misconception that the path traced by a positive test charge is a field line but actually the
path traced by a unit positive test charge represents a field full line only when it moves along a straight line.
Both the equipotential surfaces and the lines of force can be used to depict electric field in a certain region of
space. The advantage of using equipotential surfaces over the lines of force is that they give a visual picture
of both the magnitude and direction of the electric field.
Example: 42 Three positive charges of equal value q are placed at the vertices of an equilateral triangle.
The resulting lines of force should be sketched as in
(a) (b) (c) (d)
Solution (c) Option (a) shows lines of force starting from one positive charge and terminating at
another. Option (b) has one line of force making closed loop. Option (d) shows all lines
making closed loops. All these are not correct. Hence option (c) is correct
–
+
+
+
Pair of two equal and opposite
charges
Pair of two equal and similar
charges
Examples based on electric lines of
force
36
Electrostatics
Example: 43 A metallic sphere is placed in a uniform electric field. The lines of force follow the path (s)
shown in the figure as
(a) 1 (b) 2 (c) 3 (d) 4
Solution: (d) The field is zero inside a conductor and hence lines of force cannot exist inside it. Also, due
to induced charges on its surface the field is distorted close to its surface and a line of force
must deviate near the surface outside the sphere.
Example: 44 The figure shows some of the electric field lines corresponding to an electric field. The
figure suggests
[MP PMT 1999]
(a) CBA EEE (b) CBA EEE (c) BCA EEE (d) BCA EEE
Solution: (c)
Example: 45 The lines of force of the electric field due to two charges q and Q are sketched in the figure.
State if
(a) Q is positive and qQ
(b) Q is negative and qQ
(c) q is positive and qQ
(d) q is negative and qQ
Solution: (c) q is +ve because lines of force emerge from it and qQ because more lines emerge from
q and less lines terminate at Q.
Example: 46 The figure shows the lines of constant potential in a region in which an electric field is
present. The magnitude of electric field is maximum at
+
q
+
q
+
q
1
2
3
4
1
2
3
4
A
B
C
Q
q
Electrostatics
37
(a) A (b) B (c) C (d) Equal at A, B and C
Solution: (b) Since lines of force are denser at B hence electric field is maximum at B
Example: 47 Some equipotential surface are shown in the figure. The magnitude and direction of the
electric field is
(a) 100 V/m making angle 120o with the x-axis (b) 100 V/m making angle 60o with the x-
axis
(c) 200 V/m making angle 120o with the x-axis (d) None of the above
Solution: (c) By using
cosdrEdV suppose we consider line 1 and line 2 then
(30 – 20) = E cos 60o (20 – 10) × 10–2
So mvoltE /200 making in angle 120o with x-axis
Which of the following maps cannot represent an electric field
(a) (b) (c) (d)
30
V
20
V
C
B
40
V
50
V
A
10
20
30
40
30
o
20
V
30
V
40
V
E
dr
1
2
120
o
x
y
10
20
30
20
V
30
V
40
V
cm
= 30
o
y
x
Tricky example: 5
c
d
a
b
Tricky example: 6
38
Electrostatics
Solution:
(a)
If we consider a rectangular closed path, two parallel sides of it considering with
lines of force
as shown, then we find that work done along the closed path
abcd is abE1 – cdE2 0. Hence the field cannot represent a
conservative field. But electric field is a conservative field.
Hence a field represented by these lines cannot be an electric
field.
A charge Q is fixed at a distance d
in front of an infinite metal plate. The lines of force
are represented by
(a) (b) (c) (d)
Solution: (a)
Metal plate acts as an equipotential surface, therefore the field lines should act
normal to the surface of the metal plate.
Relation Between Electric Field and Potential.
In an electric field rate of change of potential with distance is known as potential
gradient. It is a vector quantity and it’s direction is opposite to
that of electric field. Potential gradient relates with electric
field according to the following relation ;
dr
dV
E This relation
gives another unit of electric field is meter
volt . In the above
relation negative sign indicates that in the direction of electric
field potential decreases.
In space around a charge distribution we can also write kEjEiEE zyx ˆ
ˆˆ
where ,
dx
dV
Ex dy
dV
Ey and
dz
dV
Ez
With the help of formula ,
dr
dV
E potential difference between any two points in an
electric field can be determined by knowing the boundary conditions
2
1
2
1
cos.. r
r
r
rdrEdrEdV
.
+
Q
A
B
dr
E
Electrostatics
39
For example: Suppose A, B and C are three points in an uniform electric field as shown in
figure.
(i) Potential difference between point A and B is
B
A
AB drEVV
Since displacement is in the direction of electric field, hence
=
0o
So, B
A
B
A
AB EddrEdrEVV 0cos
In general we can say that in an uniform electric field
d
V
E or
d
V
E|
|
Another example d
V
E
(ii) Potential difference between points A and C is :
)(cos)(cos ABEACEdrEVV C
A
AC
= – Ed
Above relation proves that potential difference between A and B is equal to the potential
difference between A and C i.e. points B and C are at same potential.
Concept
Negative of the slope of the V-r graph denotes intensity of electric field i.e. E
r
V
tan
Example: 48 The electric field, at a distance of 20 cm from the centre of a dielectric sphere of radius 10
cm is 100 V/m. The ‘E’ at 3 cm distance from the centre of sphere is
(a) 100 V/m (b) 125 V/m (c) 120 V/m (d) Zero
Solution: (c) For dielectric sphere i.e. for non-conducting sphere 2
.
r
qk
Eout and 3
R
kqr
Ein
22 )1020(
100
KQ
Eout KQ = 100 (0.2)2 so 32
222
)1010(
)103()2.0(100
in
E= 120 V/m
C
B
A
d
+
+
+
+
+
+
–
–
–
–
–
–
E
V
1
=
V
V
2
= 0
d
+
–
E
=
V/d
Example based on E =
–
dV/dr
40
Electrostatics
Example: 49 In x-y co-ordinate system if potential at a point P(x, y) is given by axyV ; where a is a
constant, if r is the distance of point P from origin then electric field at P is proportional
to [RPMT 2000]
(a) r (b) r–1 (c) r—2 (d) r2
Solution: (a) By using
dr
dV
E ay
dx
dV
Ex , ax
dy
dV
Ey
Electric field at point P aryxaEEE yx 2222 i.e., E r
Example: 50 The electric potential V at any point x, y, z (all in metres) in space is given by V = 4x2 volt.
The electric field at the point (1m, 0, 2m) in volt/metre is
(a) 8 along negative X-axis (b) 8 along positive X-axis
(c) 16 along negative X-axis (d) 16 along positive Z-
axis
Solution: (a) By using
dx
dV
E xx
dx
d
E8)4( 2 . Hence at point (1m, 0, 2m). E = – 8 volt/m i.e. 8 along
– ve x-axis.
Example: 51 The electric potential V is given as a function of distance x (metre) by V = (5x2 + 10x – 9) volt.
Value of electric field at x = 1m is
(a) – 20 V/m (b) 6 V/m (c) 11 V/m (d) – 23 V/m
Solution: (a) By using
dx
dV
E ; )1010()9105( 2 xxx
dx
d
E,
at x = 1m mVE /20
Example: 52 A uniform electric field having a magnitude E0 and direction along the positive X-axis
exists. If the electric potential V, is zero at X = 0, then, its value at X = +x will be
(a) V(x)= +xE0 (b) V(x)= – xE0 (c) V(x)= x2E0 (d) V(x)= – x2E0
Solution: (b) By using )(
)(
12
12
rr
VV
r
V
E
;
0
}0)({
0
x
xV
E V(x) = – xE0
Example: 53 If the potential function is given by V = 4x + 3y, then the magnitude of electric field
intensity at the point (2, 1) will be
(a) 11 (b) 5 (c) 7 (d) 1
Solution: (b) By using i.e., 22
yx EEE ; 4)34( yx
dx
d
dx
dV
Ex
and 3)34( yx
dy
d
dy
dV
Ey
CNE /5)3()4( 22
Electrostatics
41
The variation of potential with distance R from a fixed
point is as shown below. The
electric field at mR 5 is
(a) 2.5 volt/m
(b) – 2.5 volt/m
(c) mvolt /
5
2
(d) mvolt /
5
2
Solution: (a) Intensity at 5 m is same as at any point between B and C because the slope of BC
is
same throughout (i.e. electric field between B and C
is uniform). Therefore electric
field at R = 5m is equal to the slope of line BC hence by
dr
dV
E
;
m
V
E5.2
4
6
)50(
Note
: At R = 1m , m
V
E5.2
)02(
)05(
and at
m
R
3
potential is constant so E = 0.
Work Done in Displacing a Charge.
(1) Definition : If a charge Q displaced from one point to another point in electric field
then work done in this process is VQW where V = Potential difference between the two
position of charge Q. (
cos. rErEV where
is the angle between direction of electric
field and direction of motion of charge).
(2) Work done in terms of rectangular component of E and r : If charge Q is given a
displacement )
ˆ
ˆˆ
(321 krjrirr in an electric field ).
ˆ
ˆˆ
(321 kEjEiEE The work done is
)().( 332211 rErErEQrEQW .
Conservation of Electric Field.
As electric field is conservation, work done and hence potential difference between two
point is path independent and depends only on the position of points between. Which the charge
is moved.
5
4
3
2
1
1
2
3
4
5
6
0
Potential in volts
Distance
R
in
metres
5
4
3
2
1
1
2
3
4
5
6
O
A
B
C
Distance
R
in
metres
Potential in volts
A
B
I
II
III
W
I
=
W
II
=
W
III
Tricky example: 7
42
Electrostatics
Concept
No work is done in moving a charge on an equipotential surface.
Example: 54 A charge (– q) and another charge (+Q) are kept at two points A and B respectively.
Keeping the charge (+Q) fixed at B, the charge (– q) at A is moved to another point C such
that ABC forms an equilateral triangle of side l. The network done in moving the charge (–
q) is [MP PET 2001]
(a) l
Qq
0
4
1

(b) 2
0
4
1
l
Qq

(c) Qql
0
4
1

(d) Zero
Solution: (d) Since
l
kQ
VV CA
so 0)( AC VVqW
Example: 55 The work done in bringing a 20 coulomb charge from point A to point B for distance 0.2 m is
2 Joule. The potential difference between the two points will be (in volt)
(a) 0.2 (b) 8 (c) 0.1 (d) 0.4
Solution: (c) VQW . 2 = 20 V V = 0.1 volt
Example: 56 A charge +q is revolving around a stationary +Q in a circle of radius r. If the force between
charges is F then the work done of this motion will be
[CPMT 1975, 90, 91, 97; NCERT 1980, 83; EAMCET 1994; MP PET 1993, 95;
MNR 1998; AIIMS 1997; DCE 1995; RPET 1998]
(a) F × r (b) rF
2 (c)
r
F
2
(d) 0
Solution: (d) Since +q charge is moving on an equipotential surface so work done is zero.
Example: 57 Four equal charge Q are placed at the four corners of a body of side ‘a’ each. Work done in
removing a charge – Q from its centre to infinity is
W
I
=
W
II
=
W
I
II
III
A
B
+
Q
+
q
Ex
amples based on work
done
A
B
C
l
l
l
+
Q
–
q
Electrostatics
43
(a) 0 (b) a
Q
0
2
4
2

(c) a
Q
0
2
2

(d) a
Q
0
2
2

Solution: (c) We know that work done in moving a charge is W = QV
Here )( 0
VVQW 0
V W = Q × V0
Also a
Q
a
Q
a
Q
V
000
0
2
4
24
2/
.
4
1
4

So, a
Q
W
0
2
2

Example: 58 Two point charge 100
C and 5
C are placed at point A and B respectively with AB = 40 cm.
The work done by external force in displacing the charge 5
C from B to C, where BC = 30
cm, angle 229
0
/109
4
1
and
2CNmABC

[MP PMT 1997]
(a) 9 J (b) J
20
81 (c) J
25
9 (d) J
4
9
Solution: (d) Potential at B due to +100
C charge is
voltVB
6
2
6
910
4
9
1040
10100
109
Potential at C due to +100
C charge is
voltVC
6
2
6
910
5
9
1050
10100
109
Hence work done in moving charge +5
C from B to C
)(105 6
BC VVW
666 10
4
9
10
5
9
105W J
4
9
Example: 59 There is an electric field E in x-direction. If the work done in moving a charge 0.2 C
through a distance of 2 metres along a line making an angle 60o with the x-axis is 4J,
what is the value of E [CBSE 1995]
(a) 4 N/C (b) 8 N/C (c) CN/3 (d) 20 N/C
Solution: (d) By using VqW and
cosrEV
So,
cosrqEW
60cos22.04 EjW
E = 20 N/C
Example: 60 An electric charge of 20
C is situated at the origin of X-Y co-ordinate system. The potential
difference between the points. (5a, 0) and (– 3a, 4a) will be
(a) a (b) 2a (c) Zero (d)
2
a
Solution: (c)
a
kQ
VA
5
and
a
kQ
VB
5
0 BA VV
O
x
60
o
2
m
0.2
C
A
Q
5
a
5
a
B
(
–
3
a
,
4
a
)
A
(5
a
, 0)
Q
Q
–
Q
A
B
C
D
Q
Q
O
a
a
a
a
A
C
50
cm
30
cm
40
cm
+ 100
C
+ 50
C
/ 2
B
44
Electrostatics
Example: 61 Two identical thin rings each of radius R, are coaxially placed a distance R apart. If Q1 and
Q2 are respectively the charges uniformly spread on the two rings, the work done in moving
a charge q from the centre of one ring to that of the other is
(a) Zero (b)
24
)12)((
0
21
R
QQq

(c) R
QQq
0
21
4
2)(

(d)
24
)12(
0
2
1
R
Q
Q
q

Solution: (b) Potential at the centre of first ring 22
0
2
0
1
4
4RR
Q
R
Q
VA


Potential at the centre of second ring 22
0
1
0
2
4
4RR
Q
R
Q
VB


Potential difference between the two centres
24
))(12(
0
21
R
QQ
VV BA

Work done
24
))(12(
0
21
R
QQq
W

A point charge q moves from point A to point D along the path ABCD
in a uniform
electric field. If the co-ordinates of the points A, B, C and D are (a, b, 0), (2a, 0, 0), (a
,
– b, 0) and (0, 0, 0) then the work done by the electric field in this process will be
(a) – qEa
(b) Zero
(c) 2E (a + b)q
(d)
b
qEa
2
Solution: (a) As electric field is a conservative field
Hence the work done does not depend on path
AODABCD WW ODAO WW
= Fb cos 90o + Fa cos 180o = 0 + qEa (– 1)= – qEa
Equilibrium of Charge.
R
Q
2
Q
1
R
R
A
B
1
2
Y
X
C
B
D
A
E
X
Y
C
B
D
E
A
(a,b,0)
a
2
+
b
2
a
2
+
b
2
O
a
b
b
a
2
+
b
2
Tricky example: 8
Electrostatics
45
(1) Definition : A charge is said to be in equilibrium, if net force acting on it is zero. A
system of charges is said to be in equilibrium if each charge is separately in equilibrium.
(2) Type of equilibrium : Equilibrium can be divided in following type:
(i) Stable equilibrium : After displacing a charged particle from it's equilibrium position,
if it returns back then it is said to be in stable equilibrium. If U is the potential energy then in
case of stable equilibrium 2
2
dx
Ud is positive i.e., U is minimum.
(ii) Unstable equilibrium : After displacing a charged particle from it's equilibrium
position, if it never returns back then it is said to be in unstable equilibrium and in unstable
equilibrium 2
2
dx
Ud is negative i.e., U is maximum.
(iii) Neutral equilibrium : After displacing a charged particle from it's equilibrium
position if it neither comes back, nor moves away but remains in the position in which it was
kept it is said to be in neutral equilibrium and in neutral equilibrium 2
2
dx
Ud is zero i.e., U is
constant
(3) Guidelines to check the equilibrium
(i) Identify the charge for which equilibrium is to be analysed.
(ii) Check, how many forces acting on that particular charge.
(iii) There should be atleast two forces acts oppositely on that charge.
(iv) If magnitude of these forces are equal then charge is said to be in equilibrium then identify the
nature of equilibrium.
(v) If all the charges of system are in equilibrium then system is said to be in equilibrium
(4) Different cases of equilibrium of charge
Case – 1 : Suppose three similar charge
qQ ,
1 and 2
Q are placed along a straight
line as shown below
Charge q will be in equilibrium if
|||
|
21 FF
i.e.,
2
2
1
2
1
x
x
Q
Q ; This is the condition of
Case – 2 : Two similar charge 1
Q and 2
Q
are placed along a straight line at a
distance x from each other and a third
dissimilar charge q is placed in between
them as shown below
Charge q will be in equilibrium if
|||
|
21 FF
i.e.,
2
2
1
2
1
x
x
Q
Q .
x
x
1
x
2
q
Q
2
A
B
O
F
1
F
2
Q
1
x
x
1
x
2
q
Q
1
Q
2
A
B
O
F
2
F
1
46
Electrostatics
equilibrium of charge
q
. After following
the guidelines we can say that charge q is
in stable equilibrium and this system is not
in equilibrium
Note
:
12
11/QQ
x
x
and
21
21/QQ
x
x
e.g. if two charges +4
C and +16
C are
separated by a distance of 30 cm from each
other then for equilibrium a third charge
should be placed between them at a
distance cmx 10
4/161
30
1
or
cmx 20
2
Note
: Same short trick can be used here
to find the position of charge q as we
discussed in Case–1 i.e.,
12
11/QQ
x
x
and
21
21/QQ
x
x
It is very important to know that
magnitude of charge q can be determined if
one of the extreme charge (either 1
Q or
)
2
Q is in equilibrium i.e. if 2
Q is in
equilibrium then
2
2
1
|
|
x
x
Qq and if 1
Q is
in equilibrium then
2
1
2
|
|
x
x
Qq (It should
be remember that sign of q is opposite to that
of )or( 21 QQ )
Case
–
3
:
Two dissimilar charge
1
Q
and
2
Q
are placed along a straight line at a
distance x from each other, a third charge q should be placed out side the line
joining 1
Q and 2
Q for it to experience zero net force.
(Let |Q2| < |Q1|)
Short Trick :
For it's equilibrium. Charge q lies on the side of chare which is smallest in magnitude
and
1
21
/QQ
x
d
(5) Equilibrium of suspended charge in an electric field
(i) Freely suspended charged particle : To suspend a charged a particle freely in air under
the influence of electric field it’s downward weight should be balanced by upward electric force
for example if a positive charge is suspended freely in an electric field as shown then
x
d
Q
1
–
Q
2
q
+
Q
F
=
QE
mg
E
E
+
Q
F
=
QE
mg
+
+
+
+
+
+
+
+
+
–
–
–
–
–
–
–
–
–
F = QE
mg
+
Q
d
V
V
mgd
E
mg
neQ
or
or
Electrostatics
47
In equilibrium mgQE Q
mg
E
Note
: In the above case if direction of electric field is suddenly reversed in any figure
then acceleration of charge particle at that instant will be a = 2g.
(ii) Charged particle suspended by a massless insulated string (like simple pendulum) :
Consider a charged particle (like Bob) of mass m, having charge Q is suspended in an electric
field as shown under the influence of electric field. It turned through an angle (say
) and comes
in equilibrium.
So, in the position of equilibrium (O
position)
QET
sin ….(i)
mgT
cos ….(ii)
By squaring and adding equation (i) and (ii)
22 mgQET
Dividing equation (i) by (ii) mg
QE
tan
mg
QE
θ1
tan
(iii) Equilibrium of suspended point charge system : Suppose two small balls having
charge +Q on each are suspended by two strings of equal length l. Then for equilibrium position
as shown in figure.
e
FT
sin ….(i)
mgT
cos ….(ii)
22
2mgFT e
and mg
Fe
tan ; here 2
2
0
4
1
x
Q
Fe
and
sin
2
l
x
(iv) Equilibrium of suspended point charge system in a liquid : In the previous
discussion if point charge system is taken into a liquid of density
such that
remain same
then
In equilibrium
sin'TFe'
and
cos)( 'TgVmg
2
0
2
)(4
)(
tan xgVmgK
Q
gVmg
Fe'
QE
E
l
mg
T
sin
T
cos
T
O
O
T
sin
T
cos
T
F
e
mg
+
Q
l
l
x
+
Q
T
sin
T
cos
T
F
e
(
mg
–
V
g
)
+
Q
l
l
x
+
Q
48
Electrostatics
When this system was in air 2
0
2
4
tan mgx
Q
mg
Fe
So equating these two gives us
m
VVm
m
K
Vmkm 1
1
)(
11
If
is the density of material of ball then
σ
ρ
K
1
1
Example: 62 A charge q is placed at the centre of the line joining two equal charges Q. The system of the
three charges will be in equilibrium. If q is equal to
[CPMT 1999; MP PET 1999, MP PMT 1999; CBSE 1995; Bihar MEE 1995; IIT 1987]
(a)
2
Q
(b)
4
Q
(c)
4
Q
(d)
2
Q
Solution: (b) By using Tricky formula
2
2/
x
x
Qq
4
Q
q since q should be negative so
4
Q
q .
QE
mg
Q
Examples based on equilibrium of
charge
46 Electrostatics
Example: 63 Two point charges +4q and +q are placed at a distance L apart. A third charge Q is so placed
that all the three charges are in equilibrium. Then location and magnitude of third charge
will be [IIT-JEE 1975]
(a) At a distance
3
L
from +4q charge,
9
4q
(b) At a distance
3
L
from +4q charge,
9
4q
(c) At a distance
3
2L
from +4q charge,
9
4q
(d) At a distance
3
2L
from +q charge,
9
4q
Solution: (c) Let third charge be placed at a distance
1
x
from +4q charge as shown
Now
q
q
L
x
4
1
1
3
2L
3
2L
x
For equilibrium of q,
9
43/
42q
L
L
qQ
9
4q
Q
.
Example: 64 A drop of
6
10
kg water carries
6
10
C charge. What electric field should be applied to
balance it’s weight (assume g = 10 m/sec2) [MP PET 2002]
(a)
,/10 mV
Upward (b)
,/10 mV
Downward (c) 0.1 V/m Downward (d)
,/1.0 mV
Upward
Solution: (a) In equilibrium QE = mg
Q
mg
E
=
6
6
10
1010
= 10 V/m; Since charge is positive so electric field will be upward.
Example: 65 A charged water drop of radii 0.1
m
is under equilibrium in some electric field. The charge
on the drop is equivalent to electronic charge. The intensity of electric field is [RPET 1997]
(a)
CN /61.1
(b)
CN /2.25
(c)
CN /262
(d)
CN /1610
Solution: (c) In equilibrium QE = mg ;
Q
gr
Q
mg
E
.
3
43
19
336
106.1
1010)101.0()14.3(
3
4
= 262 N/C
Example: 66 The bob of a pendulum of mass 8
g
carries an electric charge of
10
102.39
coulomb in an
electric field of
metervolt /1020 3
and it is at rest. The angle made by the pendulum with the
vertical will be
(a) 27o (b) 45o (c) 87o (d) 127o
Solution: (b)
qET
sin
,
mgT
cos
mg
qE
tan
L
x1
x2
Q
+4q
+q
E
T cos
qE
mg
T sin
T
Electrostatics 47
1
8.9108
1020102.39
tan 6
310
o
45
Example: 67 Two small spherical balls each carrying a charge Q = 10
C
(10 micro-coulomb) are
suspended by two insulating threads of equal lengths 1 m each, from a point fixed in the
ceiling. It is found that in equilibrium threads are separated by an angle 60o between them,
as shown in the figure. What is the tension in the threads. (Given :
29
0/109
4
1CNm

) [MP PET 2001]
(a) 18 N
(b) 1.8 N
(c) 0.18 N
(d) None of these
Solution: (b) From the geometry of figure
r = 1m
In the condition of equilibrium
e
oFT 30sin
2
26
9
1
)1010(
.109
2
1
T
T= 1.8 N
Example: 68 Two similar balloons filled with helium gas are tied to L m long strings. A body of mass m is
tied to another ends of the strings. The balloons float on air at distance r. If the amount of
charge on the balloons is same then the magnitude of charge on each balloon will be
(a)
2/1
2tan
2
k
mgr
(b)
2/1
2tan
2
mgr
k
(c)
2/1
cot
2
k
mgr
(d)
2/1
tan
2
mgr
k
Solution: (a) In equilibrium
mgR2
…. (i)
sinTFe
…. (ii)
cosTR
…. (iii)
From equation (i) and (iii)
mgT
cos2
…. (iv)
Dividing equation (ii) by equation (iv)
T cos
m
mg
L
L
T sin
Fe
R
R
r
T
60
o
Q
Q
+10
C
Fe
1m
1m
30o
r
T sin
30o
T cos 30o
T
mg
30o
+10
C
30o
m
L
L
r
Q
Q
48 Electrostatics
mg
Fe
tan
2
1
mg
r
Q
k2
2
tan
2
1
2/1
2tan
2
k
mgr
Time Period of Oscillation of a Charged Body.
(1) Simple pendulum based : If a simple pendulum having length l and mass of bob m
oscillates about it's mean position than it's time period of oscillation
g
l
Tπ2
Case – 1 : If some charge say +Q is given to bob
and an electric field E is applied in the direction
as shown in figure then equilibrium position of
charged bob (point charge) changes from O to
O.
On displacing the bob from it’s equilibrium
position 0. It will oscillate under the effective
acceleration g, where
22 QEmgmg'
2
2/mQEgg'
Hence the new time period is
g'
l
T
2
1
2
1
2
2
12
QE/mg
l
T
π
Since
g'
>g, hence T1 < T
Case – 2 : If electric field is applied in the
downward direction then.
Effective acceleration
mQEgg' /
So new time period
QE/mg
l
T
π2
2
T2 < T
Case – 3 : In case 2 if electric field is applied
in upward direction then, effective acceleration.
mQEgg' /
So new time period
QE/mg
l
T
π2
3
T3 > T
Case – 4 : In the case 3,
if
2
3T
T
i.e.,
mQEg
l
/
2
g
l
2
2
1
QE = 3 mg
i.e., effective vertical force (gravity + electric)
l
O
QE
mg
mg
O
E
d
l
O
mg + QE
T
l
E
mg
l
QE
E
mg
QE
O
E
Electrostatics 49
i.e. time period of pendulum will decrease.
on the bob = mg – 3 mg = – 2 mg, hence the
equilibrium position O of the bob will be above
the point of suspension and bob will oscillate
under on effective acceleration 2g directed
upward.
Hence new time period
g
l
T2
2
4π
, T4 < T
(2) Charged circular ring : A thin stationary ring of radius R has a positive charge +Q unit.
If a negative charge – q (mass m) is placed at a small distance x from the centre. Then motion of
the particle will be simple harmonic motion.
Electric field at the location of – q charge
2
3
22
0
.
4
1
Rx
Qx
E

Since x<< R, So
2
x
neglected hence
3
0
.
4
1
R
Qx
E

Force experienced by charge – q is
3
0
.
4
1
R
Qx
qF

xF
hence motion is simple harmonic
Having time period
qQ
mRπε
πT03
4
2
(3) Spring mass system : A block of mass m containing a negative charge – Q is placed on a
frictionless horizontal table and is connected to a wall through an
unstretched spring of spring constant k as shown. If electric field E
applied as shown in figure the block experiences an electric force,
hence spring compress and block comes in new position. This is called
the equilibrium position of block under the influence of electric field.
If block compressed further or stretched, it execute oscillation having
time period
k
m
πT2
. Maximum compression in the spring due to
electric field =
k
QE
Neutral Point.
A neutral point is a point where resultant electrical field is zero. It is obtained where two
electrical field are equal and opposite. Thus neutral points can be obtained only at those points
where the resultant field is subtractive. Thus it can be obtained.
(1) At an internal point along the line joining two
like charges (Due to a system of two like point
charge) : Suppose two like charges.
1
Q
and
2
Q
are
separated by a distance x from each other along a line
as shown in following figure.
(2) At an external point along the line joining two
like charges (Due to a system of two unlike point
charge) : Suppose two unlike charge
1
Q
and
2
Q
separated by a distance x from each other.
x
x1
x2
Q1
N
Q2
l
N
Q1
Q2
x
x
O
R
– q
+Q
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
m, – Q
E
k
50 Electrostatics
If N is the neutral point at a distance
1
x
from
1
Q
and at
a distance
12 xxx
from
2
Q
then –
At N
||| 21 toduetodue QE.F.QE.F.|
i.e.,
2
1
1
0.
4
1
x
Q

=
2
2
2
0.
4
1
x
Q

2
2
1
2
1
x
x
Q
Q
Short rick :
12
11/QQ
x
x
and
21
21/QQ
x
x
Note
: In the above formula if
21 QQ
, neutral
point lies at the centre so remember that
resultant field at the midpoint of two equal
and like charges is zero.
Here neutral point lies outside the line joining two unlike
charges and also it lies nearer to charge which is smaller
in magnitude.
If
21 QQ
then neutral point will be obtained on the
side of
1
Q
, suppose it is at a distance l from
1
Q
Hence at neutral point ;
2
2
21
lx
kQ
l
kQ
2
2
1
lx
l
Q
Q
Short rick :
1
12
/QQ
x
l
Note
: In the above discussion if
|||| 21 QQ
neutral point will be at infinity.
Zero Potential Due to a System of Two Point Charge.
If both charges are like then resultant potential is not zero at any finite point because
potentials due to like charges will have same sign and can therefore never add up to zero. Such
a point can be therefore obtained only at infinity.
If the charges are unequal and unlike then all such points where resultant potential is zero
lies on a closed curve, but we are interested only in those points where potential is zero along
the line joining the two charges.
Two such points exist, one lies inside and one lies outside the charges on the line joining
the charges. Both the above points lie nearer the smaller charge, as potential created by the
charge larger in magnitude will become equal to the potential created by smaller charge at the
desired point at larger distance from it.
I. For internal point :
(It is assumed that
21 QQ
).
1
2
1
1xx
Q
x
Q
1
12
1
/QQ
x
x
II. For External point :
1
2
1
1xx
Q
x
Q
1
12
1
/QQ
x
x
Example: 69 Two similar charges of +Q as shown in figure are placed at points A and B. – q charge is
placed at point C midway between A and B. – q charge will oscillate if [RPET 1988]
(a) It is moved towards A
x
x1
x2
Q1
P
Q2
x1
Q1
Q2
x
P
D
C
– q
A
B
N
Examples based on oscillation of charge and neutral
point
Electrostatics 51
(b) It is moved towards B
(c) It is moved along CD
(d) Distance between A and B is reduced
Solution: (c) When – q charge displaced along CD, a restoring force act on it which causes oscillation.
Example: 70 Two point charges (+Q) and (– 2Q) are fixed on the X-axis at positions a and 2a from origin
respectively. At what position on the axis, the resultant electric field is zero [MP PET 2001]
(a) Only
ax 2
(b) Only
ax 2
(c) Both
ax 2
(d)
2
3a
x
only
Solution: (b) Let the electric field is zero at a point P distance d from the charge +Q so at P.
0
)(
)2(.
22
da
Qk
d
Qk
22 )(
21
dad
)12(
a
d
Since d > a i.e. point P must lies on negative x-axis as shown at a distance x from origin
hence
aa
a
adx 2
12
. Actually P lies on negative x-axis so
ax 2
.
Example: 71 Two charges 9e and 3e are placed at a distance r. The distance of the point where the
electric field intensity will be zero is [MP PMT 1989]
(a)
13
r
from 9e charge (b)
311
r
from 9e
charge
(c)
31
r
from 3e charge (d)
311
r
from 3e
charge
Solution: (b) Suppose neutral point is obtained at a distance
1
x
from charge 9e and
2
x
from charge 3e
By using
1
2
1
1Q
Q
x
x
=
e
e
r
9
3
1
3
1
1
r
Example: 72 Two point charges – Q and 2Q are separated by a distance R, neutral point will be obtained
at
(a) A distance of
)12(
R
from – Q charge and lies between the charges.
(b) A distance of
)12(
R
from – Q charge on the left side of it
(c) A distance of
)12(
R
from 2Q charge on the right side of it
+Q
+Q
P
+ Q
– 2Q
x
d
a
2a
r
x1
x2
N
9e
3e
52 Electrostatics
(d) A point on the line which passes perpendicularly through the centre of the line joining –
Q and 2Q charge.
Solution: (b) As already we discussed neutral point will be obtained on the side of charge which is
smaller in magnitude i.e. it will obtained on the left side of – Q charge and at a distance.
1
2
Q
Q
R
l
)12(
R
l
Example: 73 A charge of + 4
C is kept at a distance of 50 cm from a charge of – 6
C. Find the two points
where the potential is zero
(a) Internal point lies at a distance of 20 cm from 4
C charge and external point lies at a
distance of 100 cm from 4
C charge.
(b) Internal point lies at a distance of 30 cm from 4
C charge and external point lies at a
distance of 100 cm from 4
C charge
(c) Potential is zero only at 20 cm from 4
C charge between the two charges
(d) Potential is zero only at 20 cm from – 6
C charge between the two charges
Solution: (a) For internal point X,
cm
Q
Q
x
x20
1
4
650
1
1
2
1
and for external point Y,
cm
Q
Q
x
x100
1
4
650
1
1
2
1
Two equal negative charges – q are fixed at points (0, a) and (0, – a) on the y-axis. A
positive charge Q is released from rest at the point (2a, 0) on the x-axis. The charge Q
will
[IIT-JEE 1984, Bihar MEE 1995, MP PMT 1996]
(a) Execute simple harmonic motion about the origin
(b) Move to the origin and remains at rest
(c) Move to infinity
(d) Execute oscillatory but not simple harmonic motion.
Solution: (d) By symmetry of problem the components of force on Q due to charges at A and B along
y-axis will cancel each other while along x-axis will add up and will be along CO.
Under the action of this force charge Q will move towards O. If at any time charge Q
is at a distance x from O.
212222
0)()(
4
1
2cos2 xa
x
xa
qQ
FF

i.e.,
23
22
0
2
.
4
1
xa
qQx
F

As the restoring force F is not linear, motion will be
oscillatory (with amplitude 2a) but not simple
50c
m
100cm
4
C
X
Y
– 6
C
20cm
2a
x
a
a
O
C
A
B
– q
Q
– q
Tricky example: 9
Electrostatics 53
harmonic.
Electric Potential Energy.
(1) Potential energy of a charge : Work done in bringing the given charge from infinity to
a point in the electric field is known as potential energy of the charge. Potential can also be
written as potential energy per unit charge. i.e.
Q
U
Q
W
V
.
(2) Potential energy of a system of two charges : Since work done in bringing charge Q2
from to point B is
,
2B
VQW
where VB is potential of point B due to charge Q1 i.e.
r
Q
VB1
0
4
1

So,
r
QQ
UW 21
0
2.
4
1

This is the potential energy of charge Q2, similarly potential energy of charge Q1 will be
r
QQ
U21
0
1.
4
1

Hence potential energy of Q1 = Potential energy of Q2 = potential energy of system
r
QQ
kU 21
(in
C.G.S.
r
QQ
U21
)
Note
: Electric potential energy is a scalar quantity so in the above formula take sign of Q1
and Q2.
(3) Potential energy of a system of n charges : In a system of n charges electric potential
energy is calculated for each pair and then all energies so obtained are added algebraically. i.e.
.........
4
1
23
32
12
21
0r
QQ
r
QQ
U

and in case of continuous distribution of charge. As
VdQdU .
dQVU
e.g. Electric potential energy for a system of three charges
Potential energy
31
13
23
32
12
21
0
4
1
r
QQ
r
QQ
r
QQ

While potential energy of any of the charge say Q1 is
31
13
12
21
0
4
1
r
QQ
r
QQ

Q1
r
Q2
A
B
Q1
Q2
Q3
r31
r23
r12
54 Electrostatics
Note
: For the expression of total potential energy of a system of n charges consider
2
)1( nn
number of pair of charges.
(4) Electron volt (eV) : It is the smallest practical unit of energy used in atomic and
nuclear physics. As electron volt is defined as “the energy acquired by a particle having one
quantum of charge 1e when accelerated by 1volt” i.e.
C
J
CeV 1
106.11 19
J
19
106.1
= 1.6
10–12 erg
Energy acquired by a charged particle in eV when it is accelerated by V volt is E = (charge in
quanta) × (p.d. in volt)
Commonly asked examples :
S.No.
Charge
Accelerated by
p.d.
Gain in K.E.
(i)
Proton
5 104 V
K = e 5 104 V = 5 104 eV = 8 10–15 J [JIPMER 1999]
(ii)
Electron
100 V
K = e 100 V = 100 eV = 1.6 10–17 J [MP PMT 2000; AFMC
1999]
(iii)
Proton
1 V
K = e 1 V = 1 eV = 1.6 10–19 J [CBSE 1999]
(iv)
0.5 C
2000 V
K = 0.5 2000 = 1000 J [JIPMER 2002]
(v)
-
particle
106 V
K = (2e) 106 V = 2 MeV [MP PET/PMT 1998]
(5) Electric potential energy of a uniformly charged sphere : Consider a uniformly
charged sphere of radius R having a total charge Q. The electric potential energy of this sphere
is equal to the work done in bringing the charges from infinity to assemble the sphere.
R
Q
U
0
2
20
3

(6) Electric potential energy of a uniformly charged thin spherical shell :
R
Q
U
0
2
8

(7) Energy density : The energy stored per unit volume around a point in an electric field
is given by
2
0
2
1
Volume E
U
Ue
. If in place of vacuum some medium is present then
2
0
2
1EU re
Concepts
Electric potential energy is not localised but is distributed all over the field
If a charge moves from one position to another position in an electric field so it’s potential energy change and
Electrostatics 55
work done in this changing is
if UUW
If two similar charge comes closer potential energy of system increases while if two dissimilar charge comes
closer potential energy of system decreases.
Example: 74 If the distance of separation between two charges is increased, the electrical potential
energy of the system
[AMU 1998]
(a) May increases or decrease (b) Decreases
(c) Increase (d) Remain the same
Solution: (a) Since we know potential energy
r
QQ
kU 21
.
As r increases, U decreases in magnitude. However depending upon the fact whether both
charges are similar or disimilar, U may increase or decrease.
Example: 75 Three particles, each having a charge of 10
C are placed at the corners of an equilateral
triangle of side 10cm. The electrostatic potential energy of the system is (Given
)/109
4
1229
0CmN

[AMU 1998]
(a) Zero (b) Infinite (c) 27 J (d) 100 J
Solution: (c) Potential energy of the system,
3
1.0
)1010(
10926
9
U
= 27 J
Example: 76 Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as
shown. The net electrostatic energy of the configuration is zero if Q is equal to [IIT (Screening) 2000]
(a)
21
q
(b)
21
2
q
(c) – 2 q
(d) +q
Solution: (b) Potential energy of the configuration
0
2
.
.
.2 a
Qq
k
a
qk
a
Qq
kU
12
2
q
Q
10
C
10
C
10
C
10 cm
10 cm
10 cm
Examples based on electric potential
energy
+q
+q
Q
a
Q
q
U = 0
56 Electrostatics
Example: 77 A charge 10 e.s.u. is placed at a distance of 2 cm from a charge 40 e.s.u. and 4 cm from
another charge of 20 e.s.u. The potential energy of the charge 10 e.s.u. is (in ergs) [CPMT 1976]
(a) 87.5 (b) 112.5 (c) 150 (d) 250
Solution: (d) Potential energy of 10 e.s.u. charge is
.250
4
2010
2
4010 ergU
Example: 78 In figure are shown charges q1 = + 2 × 10–8 C and q2 = – 0.4 × 10–8 C. A charge q3 = 0.2 × 10–
8 C in moved along the arc of a circle from C to D. The potential energy of q3 [CPMT 1986]
(a) Will increase approximately by 76%
(b) Will decreases approximately by 76%
(c) Will remain same
(d) Will increases approximately by 12%
Solution: (b) Initial potential energy of q3
9
3231 109
18.0
qqqq
Ui
Final potential energy of q3
9
3231 109
2.08.0
qqqq
Uf
Change in potential energy = Uf – Ui
Now percentage change in potential energy
100
i
if
u
UU
18.0
1001
2.0
1
21
3
32
qq
q
qq
On putting the values
%76–
~
Three charged particles are initially in position 1. They are free to move and they
come in position 2 after some time. Let U1 and U2 be the electrostatics potential
energies in position 1 and 2. Then
(a) U1 > U2 (b) U2 > U1
(c) U1 = U2 (d) U2 U1
Solution: (a) Particles move in a direction where potential energy of the system is decreased.
40 esu
10 esu
20 esu
4 cm
2 cm
C
q1
q3
80
cm
D
60 cm
80 cm
A
B
q2
Tricky example: 10
q3
C
A
B
q1
q2
80 cm
D
60 cm
Electrostatics 57
Motion of Charged Particle in an Electric Field.
(1) When charged particle initially at rest is placed in the uniform field :
Let a charge particle of mass m and charge Q be initially at rest in an electric field of
strength E
(i) Force and acceleration : The force experienced by the charged particle is
QEF
.
Positive charge experiences force in the direction of electric field while negative charge
experiences force in the direction opposite to the field. [Fig. (A)]
Acceleration produced by this force is
m
QE
m
F
a
Since the field E in constant the acceleration is constant, thus motion of the particle is
uniformly accelerated.
(ii) Velocity : Suppose at point A particle is at rest and in time t, it reaches the point B [Fig.
(B)]
V = Potential difference between A and B; S = Separation between A and B
(a) By using
atuv
,
t
m
E
Qv 0
,
m
QEt
v
(b) By using
asuv 2
22
,
s
m
QE
v 20
2
m
QV
v2
2
s
V
E
m
QV
v2
(iii) Momentum : Momentum p = mv,
QEt
m
QEt
mp
or
mQV
m
QV
mp 2
2
(iv) Kinetic energy : Kinetic energy gained by the particle in time t is
m
tEQ
m
QEt
mmvK2
)(
2
1
2
12222
2
or
QV
m
QV
mK 2
2
1
(2) When a charged particle enters with an initial velocity at right angle to the uniform
field :
Fig.
(A)
E
A
B
S
Fig.
(B)
+Q
– Q
F=QE
F = QE
E
58 Electrostatics
When charged particle enters perpendicularly in an electric field, it describe a parabolic
path as shown
(i) Equation of trajectory : Throughout the motion particle has uniform velocity along x-
axis and horizontal displacement (x) is given by the equation x = ut
Since the motion of the particle is accelerated along y–axis, we will use equation of motion
for uniform acceleration to determine displacement y. From
2
2
1atutS
We have
0u
(along y-axis) so
2
2
1aty
i.e., displacement along y-axis will increase rapidly with time (since
)
2
ty
From displacement along x-axis
u
x
t
So
2
2
1
u
x
m
QE
y
; this is the equation of parabola which shows
2
xy
(ii) Velocity at any instant : At any instant t,
uvx
and
m
QEt
vy
So
2
222
222
|| m
tEQ
uvvvv yx
If
is the angle made by v with x-axis than
mu
QEt
v
v
x
y
tan
.
Concepts
An electric field is completely characterized by two physical quantities Potential and Intensity. Force
characteristic of the field is intensity and work characteristic of the field is potential.
If a charge particle (say positive) is left free in an electric field, it experiences a force
)( QEF
in the direction
of electric field and moves in the direction of electric field (which is desired by electric field), so its kinetic
energy increases, potential energy decreases, then work is done by the electric field and it is negative.
Example: 79 An electron (mass =
kg
31
101.9
and charge =
.)106.1 19 coul
is sent in an electric field of
intensity
./1016mV
How long would it take for the electron, starting from rest, to attain
one–tenth the velocity of light
(a)
sec
12
107.1
(b)
sec
6
107.1
(c)
sec
8
107.1
(d)
sec
10
107.1
Q
E
Examples based on motion of
charge
vx
vy
v
Y
X
E
u
P(x, y)
Electrostatics 59
Solution: (b) By using
m
QEt
v
31
619
8
101.9
10)106.1(
103
10
1
t
.107.1 10 sect
Example: 80 Two protons are placed
m
10
10
apart. If they are repelled, what will be the kinetic energy
of each proton at very large distance
(a)
J
19
1023
(b)
J
19
105.11
(c)
J
19
1056.2
(d)
J
28
1056.2
Solution: (d) Potential energy of the system when protons are separated by a distance of
10
10
m is
JU 19
10
21991023
10
)106.1(109
According to law of conservation of energy at very larger distance, this energy is equally
distributed in both the protons as their kinetic energy hence K.E. of each proton will be
.105.11 19 J
Example: 81 A particle A has a charge +q and particle B has charge +4q with each of them having the
same mass m. When allowed to fall from rest through the same electrical potential
difference, the ratio of their speeds
B
A
v
v
will becomes [BHU 1995; MNR 1991]
(a) 2 : 1 (b) 1 : 2 (c) 1 : 4 (d) 4 : 1
Solution: (b) We know that kinetic energy
QVmvK 2
2
1
. Since, m and V are same so,
Qv
2
.
2
1
4 q
q
Q
Q
v
v
B
A
B
A
Example: 82 How much kinetic energy will be gained by an
particle in going from a point at 70 V to
another point at 50 V [RPET 1996]
(a) 40 eV (b) 40 keV (c) 40 MeV (d) 0 eV
Solution: (a) Kinetic energy
VQK
VeK )5070()2(
eV40
Example: 83 A particle of mass 2g and charge
C
1
is held at a distance of 1 metre from a fixed charge of
.1mC
If the particle is released it will be repelled. The speed of the particle when it is at a
distance of 10 metres from the fixed charge is [CPMT 1989]
(a) 100 m/s (b) 90 m/s (c) 60 m/s (d) 45 m/s
Solution: (b) According to conservation of energy
Energy of moving charge at
A
Energy of moving charge at B
23
63
9
63
9)102(
2
1
10
1010
109
1
1010
109v
1m
10m
A
B
1 mC
1
C
Moving
charge
P+
p+
60 Electrostatics
8100
2v
m/sec90v
A mass of 1g carrying charge q falls through a potential difference V. The kinetic
energy acquired by it is E. When a mass of 2g carrying the charge q falls through a
potential difference V. What will be the kinetic energy acquired by it
(a) 0.25 E (b) 0.50 E (c) 0.75 E (d) E
Solution: (d) In electric field kinetic energy gain by the charged particle K = qV. Which depends
charge and potential difference applied but not on the mass of the charged particle.
Force on a Charged Conductor.
To find force on a charged conductor (due to repulsion of like charges) imagine a small part
XY to be cut and just separated from the rest of the conductor MLN. The field in the cavity due
to the rest of the conductor is E2, while field due to small part is E1. Then
Inside the conductor
0
21 EEE
or
21 EE
Outside the conductor
0
21
EEE
Thus
0
21 2
EE
To find force, imagine charged part XY (having charge
dA
placed in the cavity MN having
field E2). Thus force
2
)( EdAdF
or
dAdF
0
2
2
. The force per unit area or electric pressure is
0
2
2
dA
dF
The force is always outwards as
2
)(
is positive i.e., whether charged positively or
negatively, this force will try to expand the charged body.
A soap bubble or rubber balloon expands on given charge to it (charge of any kind + or –).
Equilibrium of Charged Soap Bubble.
For a charged soap bubble of radius R and surface tension T and charge density
.
The
pressure due to surface tension
R
T
4
and atmospheric pressure
out
P
act radially inwards and the
electrical pressure
)( el
P
acts radially outward.
The total pressure inside the soap bubble
0
2
outin 2
4
R
T
PP
Tricky example: 11
(A)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
M
N
Y
X
E2
E1
E1
E2
Inside
E = 0
L
+
+
+
+
+
+
+
+
+
+
+
+
+
E2
(B)
Electrostatics 61
Excess pressure inside the charged soap bubble
0
2
excessoutin 2
4
R
T
PPP
. If air pressure
inside and outside are assumed equal then
outin PP
i.e.,
0
excess P
. So,
0
2
2
4
R
T
This result give us the following formulae
(1) Radius of bubble
2
0
8
T
R
(2) Surface tension
0
2
8
R
T
(3) Total charge on the bubble
TRRQ 0
28
(4) Electric field intensity at the surface of the bubble
R
kT
R
T
E
328
0
(5) Electric potential at the surface
0
8
3
RT
RTkV
ai
r
Pout
Pin
PT
ai
r
Uncharged
Pout
Pin
PT
+
+
+
+
Pelec
Charged
60 Electrostatics
Electric Dipole.
(1) General information : System of two equal and opposite charges separated by a small
fixed distance is called a dipole.
(i) Dipole axis : Line joining negative charge to positive charge of a dipole is called its axis.
It may also be termed as its longitudinal axis.
(ii) Equatorial axis : Perpendicular bisector of the dipole is called its equatorial or
transverse axis as it is perpendicular to length.
(iii) Dipole length : The distance between two charges is known as dipole length (L = 2l)
(iv) Dipole moment : It is a quantity which gives information about the strength of dipole.
It is a vector quantity and is directed from negative charge to positive charge along the axis. It
is denoted as
p
and is defined as the product of the magnitude of either of the charge and the
dipole length.
i.e.
)2( lqp
Its S.I. unit is coulomb-metre or Debye (1 Debye = 3.3 × 10–30 C m) and its dimensions are
M0L1T1A1.
Note
: A region surrounding a stationary electric dipole has electric field only.
When a dielectric is placed in an electric field, its atoms or molecules are
considered as tiny dipoles.
Water (H2O), Chloroform (CHCl3), Ammonia (NH3), HCl, CO molecules are some
example of permanent electric dipole.
A
B
+q
2l
Equatorial
axis
Dipole
axis
–
q
+
+
–
–
O2–
H+
H+
+
–
+
Electrostatics 61
(2) Electric field and potential due to an electric dipole : It is better to understand
electric dipole with magnetic dipole.
S.No.
Electric dipole
Magnetic dipole
(i)
System of two equal and opposite charges
separated by a small fixed distance.
System of two equal and opposite magnetic
poles (Bar magnet) separated by a small
fixed distance.
(ii)
Electric dipole moment :
)2( lqp
, directed
from
q
to +q. It’s S.I. unit is coulomb ×
meter or Debye.
Magnetic dipole moment :
)2( lmM
,
directed from S to N. It’s S.I. unit is ampere
× meter2.
(iii)
Intensity of electric field
If a, e and g are three points on axial,
equatorial and general position at a
distance r from the centre of dipole
on axial point
3
0
2
.
4
1
r
p
Ea

(directed from – q
to +q)
on equatorial point
3
0.
4
1
r
p
Ee

(directed from +q
to –q)
on general point
)1cos3(.
4
12
3
0

r
p
Ea
Angle between –
a
E
and
p
is 0o,
e
E
and
p
is
180o,
E
and
p
is (
+
) (where
tan
2
1
tan
)
Electric Potential – At a
2
0.
4
1
r
p
Va

, At e
0V
Intensity of magnetic field
If a, e and g are three points on axial,
equatorial and general position at a distance
r from the centre of dipole
on axial point
3
02
.
4r
M
Ba
(directed from S
to N)
on equatorial point
3
0.
4r
M
Be
(directed from
N to S)
on general point
)1cos3(.
42
3
0
r
M
Ba
Angle between –
a
B
and
M
is 0o,
e
B
and
M
is 180o,
B
and
M
is (
+
) (where
tan
2
1
tan
)
A
–q
+q
2l
B
p
a
+q
Equatorial
line
Axial
line
+
a
E
g
E
e
E
e
2l
p
– q
a
Equatorial
line
Axial
line
e
+
a
B
e
B
B
M
g
2l
N
S
N
S
– m
+ m
M
62 Electrostatics
At g
2
0
cos
.
4
1
r
p
V

(3) Dipole (electric/magnetic) in uniform field (electric/magnetic)
(i) Torque : If a dipole is placed in an uniform field such that dipole (i.e.
p
or
M
) makes
an angle
with direction of field then two equal and opposite force acting on dipole constitute a
couple whose tendency is to rotate the dipole hence a torque is developed in it and dipole tries
to align it self in the direction of field.
Consider an electric dipole in placed in an
uniform electric field such that dipole (i.e.
p
)
makes an angle
with the direction of electric
field as shown
(a) Net force on electric dipole
0
net
F
(b) Produced torque
= pE sin
)( EP
A magnetic dipole of magnetic moment M is
placed in uniform magnetic field B by making
an angle
as shown
(a) Net force on magnetic dipole
0
net
F
(b) torque
= MB sin
)( BM
(ii) Work : From the above discussion it is clear that in an uniform electric/magnetic field
dipole tries to align itself in the direction of electric field (i.e. equilibrium position). To change
it’s angular position some work has to be done.
Suppose an electric/magnetic dipole is kept in an uniform electric/magnetic field by
making an angle
1 with the field, if it is again turn so that it makes an angle
2 with the field,
work done in this process is given by the formula
)cos(cos 21
pEW
If
1 = 0o and
2 =
i.e. initially dipole is
kept along the field then it turn through
)cos(cos 21
MBW
If
1 = 0o and
2 =
then W = MB(1 – cos
)
+q
+q
–q
–q
1
2
E
B
1
2
M
F =
qE
F =
qE
B
A
–q
+q
N
S
F = mB
F = mB
M
Electrostatics 63
so work done
)cos1(
pEW
(iii) Potential energy : In case of a dipole (in a uniform field), potential energy of dipole is
defined as work done in rotating a dipole from a direction perpendicular to the field to the given
direction i.e. if
1 = 90o and
2 =
then –
)cos90(cos
pEUW
U = – pE cos
)cos90(cos
MBUW
U = – MB cos
(iv) Equilibrium of dipole : We know that, for any equilibrium net torque and net force on
a particle (or system) should be zero.
We already discussed when a dipole is placed in an uniform electric/magnetic field net
force on dipole is always zero. But net torque will be zero only when
= 0o or 180o
When
= 0o i.e. dipole is placed along the electric field it is said to be in stable equilibrium,
because after turning it through a small angle, dipole tries to align itself again in the direction
of electric field.
When
= 180o i.e. dipole is placed opposite to electric field, it is said to be in unstable
equilibrium.
= 0o
= 90O
= 180o
Stable equilibrium Unstable
equilibrium
= 0
max = pE
= 0
W = 0 W = pE Wmax = 2pE
Umin = – pE U = 0 Umax = pE
= 0o
= 90O
= 180o
Stable equilibrium
Unstable equilibrium
= 0
max = MB
= 0
W = 0 W = MB Wmax = 2MB
Umin = – MB U = 0 Umax = MB
(v) Angular SHM : In a uniform electric/magnetic field (intensity E/B) if a dipole
(electric/magnetic) is slightly displaced from it’s stable equilibrium position it executes angular
SHM having period of oscillation. If I = moment of inertia of dipole about the axis passing
through it’s centre and perpendicular to it’s length.
P
P
E
M
M
B
E
p
E
E
p
M
B
M
M
B
B
64 Electrostatics
For electric dipole :
pE
I
T
2
and For Magnetic dipole :
MB
I
T
2
(vi) Dipole-point charge interaction : If a point charge/isolated magnetic pole is placed
in dipole field at a distance r from the mid point of dipole then force experienced by point
charge/pole varies according to the relation
3
1
r
F
(vii) Dipole-dipole interaction : When two dipoles placed closed to each other, they
experiences a force due to each other. If suppose two dipoles (1) and (2) are placed as shown in
figure then
Both the dipoles are placed in the field of one another hence potential energy dipole (2) is
31
0
212122 2
.
4
1
0cos r
p
pEpEpU

then by using
dr
dU
F
, Force on dipole (2) is
dr
dU
F2
2
321
0
22
.
4
1
r
pp
dr
d
F

421
0
6
.
4
1
r
pp

Similarly force experienced by dipole (1)
421
0
16
.
4
1
r
pp
F

so
421
0
21 6
.
4
1
r
pp
FF

Negative sign indicates that force is attractive.
421
0
6
.
4
1
|| r
pp
F

and
4
1
r
F
S. No.
Relative position of dipole
Force
Potential energy
(i)
421
0
6
.
4
1
r
pp

(attractive)
321
0
2
.
4
1
r
pp

(ii)
421
0
3
.
4
1
r
pp

(repulsive)
321
0
.
4
1
r
pp

(iii)
421
0
3
.
4
1
r
pp

(perpendicular to
r )
0
+q
+q
–q
– q
O
O
E2
E1
P1
P2
r
1
2
1
P
+q
– q
2
P
+q
– q
+q
+q
–q
–q
2
P
1
P
r
+q
–q
1
P
r
2
P
+q
– q
Electrostatics 65
Note
: Same result can also be obtained for magnetic dipole.
(4) Electric dipole in non-uniform electric field : When an electric dipole is placed in a
non-uniform field, the two charges of dipole experiences unequal forces, therefore the net force
on the dipole is not equal to zero. The magnitude of the force is given by the negative derivative
of the potential energy w.r.t. distance along the axis of the dipole i.e.
dr
Ed
p
dr
dU
F.
.
Due to two unequal forces, a torque is produced which rotate the
dipole so as to align it in the direction of field. When the dipole gets
aligned with the field, the torque becomes zero and then the
unbalanced force acts on the dipole and the dipole then moves
linearly along the direction of field from weaker portion of the field
to the stronger portion of the field. So in non-uniform electric field
(i) Motion of the dipole is translatory and rotatory
(ii) Torque on it may be zero.
Concepts
For a short dipole, electric field intensity at a point on the axial line is double than at a point on the
equatorial line on electric dipole i.e. Eaxial = 2Eequatorial
It is intresting to note that dipole field
3
1
r
E
decreases much rapidly as compared to the field of a point charge
.
1
2
r
E
Example: 84 If the magnitude of intensity of electric field at a distance x on axial line and at a distance y
on equatorial line on a given dipole are equal, then x : y is [EAMCET 1994]
(a) 1 : 1 (b)
2:1
(c) 1 : 2 (d)
1:2
3
Solution: (d) According to the question
3
0
3
0
.
4
12
.
4
1
y
p
x
p

1:)2( 3/1
y
x
Example: 85 Three charges of (+2q), (– q) and (– q) are placed at the corners A, B and C of an equilateral
triangle of side a as shown in the adjoining figure. Then the dipole moment of this
combination is [MP PMT 1994; CPMT 1994]
(a) qa
(b) Zero
(c)
3aq
Examples based on electric
dipole
–
q
+q
+q
– q
qE
qE
E > E
A
B
C
+2q
– q
– q
a
a
a
66 Electrostatics
(d)
qa
3
2
Solution: (c) The charge +2q can be broken in +q, +q. Now as shown in figure we have two equal dipoles
inclined at an angle of 60o. Therefore resultant dipole moment will be
60cos2
22 pppppnet
p3
qa3
Example: 86 An electric dipole is placed along the x-axis at the origin O. A point P is at a distance of 20
cm from this origin such that OP makes an angle
3
with the x-axis. If the electric field at P
makes an angle
with x-axis, the value of
would be [MP PMT 1997]
(a)
3
(b)
2
3
tan
31
(c)
3
2
(d)
2
3
tan 1
Solution: (b) According to question we can draw following figure.
As we have discussed earlier in theory
3
3
tan
2
1
tan
2
3
tan 1
So,
2
3
tan
31
Example: 87 An electric dipole in a uniform electric field experiences [RPET 2000]
(a) Force and torque both (b) Force but no torque (c) Torque but no
force (d) No force and no torque
Solution: (c) In uniform electric field Fnet = 0,
net 0
Example: 89 Two opposite and equal charges 4 × 10–8 coulomb when placed 2 × 10–2 cm away, form a
dipole. If this dipole is placed in an external electric field 4 × 108 newton/coulomb, the
value of maximum torque and the work done in rotating it through 180o will be [MP PET 1996 Similar to MP PMT 1987]
(a) 64 × 10–4 Nm and 64 × 10–4 J (b) 32 × 10–4 Nm and 32 × 10–4 J
(c) 64 × 10–4 Nm and 32 × 10–4 J (d) 32 × 10–4 Nm and 64 × 10–4 J
Solution: (d)
max = pE and Wmax = 2pE p = Q × 2l = 4 × 10–8 × 2 × 10–2 × 10–2 = 8 × 10–12 C-m
So,
max = 8 × 10–12 × 4 × 108 = 32 × 10–4 N-m and Wmax = 2 × 32 × 10–4 = 64 × 10–4 J
Example: 90 A point charge placed at any point on the axis of an electric dipole at some large distance
experiences a force F. The force acting on the point charge when it’s distance from the
dipole is doubled is
[CPMT 1991; MNR 1986]
(a) F (b)
2
F
(c)
4
F
(d)
8
F
Solution: (d) Force acting on a point charge in dipole field varies as
3
1
r
F
where r is the distance of
point charge from the centre of dipole. Hence if r makes double so new force
8
'F
F
.
60O
P
P
O
y
x
P
E
/3
p
Electrostatics 67
Example: 91 A point particle of mass M is attached to one end of a massless rigid non-conducting rod of
length L. Another point particle of the same mass is attached to other end of the rod. The
two particles carry charges +q and – q respectively. This arrangement is held in a region of
a uniform electric field E such that the rod makes a small angle
(say of about 5 degrees)
with the field direction (see figure). Will be minimum time, needed for the rod to become
parallel to the field after it is set free [CPMT 1995]
(a)
pE
mL
t2
2
(b)
qE
mL
t22
(c)
pE
mL
t22
3
(d)
qE
mL
t2
Solution: (b) In the given situation system oscillate in electric field with maximum angular displacement
.
It’s time period of oscillation (similar to dipole)
pE
I
T
2
where I = moment of inertia of the system and
qLp
Hence the minimum time needed for the rod becomes parallel to the field is
pE
IT
t24
Here
222
2
22 MLL
M
L
MI
qE
ML
EqL
ML
t2222
2
An electric dipole is placed at the origin O and is directed along the x-axis. At a point
P, far away from the dipole, the electric field is parallel to y-axis. OP makes an angle
with the x-axis then
(a)
3tan
(b)
2tan
(c)
= 45o (d)
2
1
tan
Solution: (b) As we know that in this case electric field makes an angle
+
with the direction of
dipole
Where
tan
2
1
tan
Here
+
= 90o
90
Hence
tan
2
1
)90tan(
tan
2
1
cot
2tan 2
2tan
Electric Flux.
(1) Area vector : In many cases, it is convenient to treat area of a surface as a vector. The
length of the vector represents the magnitude of the area and its direction is along the outward
drawn normal to the area.
Area ds
sd
Tricky example: 12
O
Y
X
E
P
P
+q
–q
E
68 Electrostatics
(2) Electric flux : The electric flux linked with any surface in an electric field is basically a
measure of total number of lines of forces passing normally through the surface. or
Electric flux through an elementary area
ds
is defined as the scalar product of area of field
i.e.
cosdsEdsEd
Hence flux from complete area (S)
cosdsE
= ES cos
If
= 0o, i.e. surface area is perpendicular to the electric field,
so flux linked with it will be max.
i.e.
max = E ds and if
= 90o,
min = 0
(3) Unit and Dimensional Formula
S.I. unit – (volt × m) or
2
m
CN
It’s Dimensional formula – (ML3T–3A– 1)
(4) Types : For a closed body outward flux is taken to be positive, while inward flux is to
be negative
Gauss’s Law.
(1) Definition : According to this law, total electric flux through a closed surface enclosing
a charge is
0
1
times the magnitude of the charge enclosed i.e.
)(
1.
0enc
Q
(2) Gaussian Surface : Gauss’s law is valid for symmetrical charge distribution. Gauss’s
law is very helpful in calculating electric field in those cases where electric field is symmetrical
around the source producing it. Electric field can be calculated very easily by the clever choice
of a closed surface that encloses the source charges. Such a surface is called “Gaussian surface”.
This surface should pass through the point where electric field is to be calculated and must have
a shape according to the symmetry of source.
ds
sd
E
(B)
Body
Negative-flux
(A)
Positive – flux
Body
n
E
Electrostatics 69
e.g. If suppose a charge Q is placed at the centre of a hemisphere, then to calculate the flux
through this body, to encloses the first charge we will have to imagine a Gaussian surface. This
imaginary Gaussian surface will be a hemisphere as shown.
Net flux through this closed body
0
Q
Hence flux coming out from given hemisphere is
.
20
Q
(3) Zero flux : The value of flux is zero in the following circumstances
(i) If a dipole is enclosed by a surface
0;0 enc
Q
(ii) If the magnitude of positive and
negative charges are equal inside a closed
surface
,0
enc
Q
so,
= 0
(iii) If a closed body (not enclosing any charge) is placed in an electric field (either
uniform or non-uniform) total flux linked with it will be zero
0
T
2
aE
utoin
(4) Flux emergence : Flux linked with a closed body is independent of the shape and size of
the body and position of charge inside it
Spher
e
0
T
Q
+q
–q
+q
+4q
–5q
sd
sd
a
y
E
a
x
sd
0
Q
T
Q
0
Q
T
0
Q
T
0
Q
T
Q
Q
Q
ER
in 2
ER
out 2
0
T
E
70 Electrostatics
(i) If a hemispherical body is placed in
uniform electric field then flux linked with
the curved surface
ER
curved 2
(ii) If a hemispherical body is placed in
non-uniform electric field as shown below.
then flux linked with the curved surface.
ER
curved 2
2
(v) If charge is kept at the centre of cube
).(
1
0
Q
total
0
6
Q
face
0
8
Q
corner
0
12
Q
edge
(iv) If charge is kept at the centre of a face
First we should enclosed the charge by
assuming a Gaussian surface (an identical
imaginary cube)
0
Q
total
0
2
Q
cube
(i.e. from 5 face
only)
00 1025
1
QQ
face
.
Concept
In C.G.S.
4
1
0
. Hence if 1C charge is enclosed by a closed surface so flux through the surface will be
4
.
Example: 91 Electric charge is uniformly distributed along a long straight wire of radius 1 mm. The
charge per cm length of the wire is Q coulomb. Another cylindrical surface of radius 50 cm
and length 1 m symmetrically encloses the wire as shown in the figure. The total electric
flux passing through the cylindrical surface is [MP PET 2001]
E
n
ˆ
R
n
ˆ
R
Example based on electric flux and Gauss’s
law
Q
Q
Electrostatics 71
(a)
0
Q
(b)
0
100
Q
(c)
)(
10
0

Q
(d)
)(
100
0

Q
Solution: (b) Given that charge per cm length of the wire is Q. Since 100 cm length of the wire is
enclosed so
QQenc 100
Electric flux emerging through cylindrical surface
0
100
Q
.
Example: 92 A charge Q is situated at the corner A of a cube, the electric flux through the one face of the
cube is
[CPMT 2000]
(a)
0
6
Q
(b)
0
8
Q
(c)
0
24
Q
(d)
0
2
Q
Solution: (c) For the charge at the corner, we require eight cube to symmetrically enclose it in a
Gaussian surface. The total flux
0
Q
T
. Therefore the flux through one cube will be
.
80
Q
cube
The cube has six faces and flux linked with three faces (through A) is zero, so
flux linked with remaining three faces will
.
80
Now as the remaining three are identical
so flux linked with each of the three faces will be
00 24
1
8
1
3
1
QQ
.
Example: 93 A square of side 20 cm is enclosed by a surface of sphere of 80 cm radius. Square and
sphere have the same centre. Four charges + 2 × 10–6 C, – 5 × 10– 6 C, – 3 × 10– 6 C, +6 × 10– 6
C are located at the four corners of a square, then out going total flux from spherical
surface in N–m2/C will be [RPMT 1989]
(a) Zero (b) (16
) × 10– 6 (c) (8
) × 10–6 (d) 36
× 10–6
Solution: (a) Since charge enclosed by Gaussian surface is
0)106103105102( 6666
.
enc
so
0
Example: 94 In a region of space, the electric field is in the x-direction and proportional to x, i.e.,
ixEE ˆ
0
. Consider an imaginary cubical volume of edge a, with its edges parallel to the
axes of coordinates. The charge inside this cube is
50
cm
1 m
+
+
+
+
+
+
Q
72 Electrostatics
(a) Zero (b)
3
00 aE
(c)
3
0
0
1aE
(d)
2
00
6
1aE
Solution: (b) The field at the face ABCD =
.
ˆ
00 ixE
Flux over the face ABCD = – (E0x0)a2
The negative sign arises as the field is directed into the
cube.
The field at the face EFGH =
.
ˆ
)( 00 iaxE
Flux over the face EFGH =
2
00 )( aaxE
The flux over the other four faces is zero as the field is
parallel to the surfaces.
Total flux over the cube
qaE 2
1
2
0
where q is the total charge inside the cube.
.
3
00 aEq
In the electric field due to a point charge + Q a spherical closed surface is drawn as
shown by the dotted circle. The electric flux through the surface drawn is zero by
Gauss’s law. A conducting sphere is inserted intersecting the previously drawn
Gaussian surface. The electric flux through the surface
(a) Still remains zero
(b) Non zero but positive
(c) Non-zero but negative
(d) Becomes infinite
Solution: (b) Due to induction some positive charge will lie within the Gaussian surface drawn and
hence flux becomes something positive.
Application of Gauss’s Law.
Gauss’s law is a powerful tool for calculating electric field in case of symmetrical charge
distribution by choosing a Gaussian surface in such away that
E
is either parallel or
perpendicular to it’s various faces.
e.g. Electric field due to infinitely long line of charge : Let us
consider a uniformly charged wire of infinite length having a constant
linear charge density is
.
length
charge
Let P be a point distant r from the
wire at which the electric field is to be calculated.
+
+
+
+
+
+
+
P
l
1
2
a
n
ˆ
n
ˆ
D
C
G
a
a
X
H
B
A
E
F
a
x0
Y
Z
Tricky example: 13
+Q
Electrostatics 73
Draw a cylinder (Gaussian surface) of radius r and length l around the line charge which
encloses the charge Q (
lQ .
). Cylindrical Gaussian surface has three surfaces; two circular
and one curved for surfaces (1) and (2) angle between electric field and normal to the surface is
90o i.e.,
.90 o
So flux linked with these surfaces will be zero. Hence total flux will pass through curved
surface and it is
cosdsE
…. (i)
According to Gauss’s law
0
Q
…. (ii)
Equating equation (i) and (ii)
0
Q
dsE
00
2
Q
rlEx
Q
dsE
r
k
rrl
Q
E


2
22 00
0
4
1

K